Power functions
Four times the workforce gives twice the output. Which kind of function behaves like that?
We insist on $x>0.$ The reason is that $x^{r}$ simply is not defined for negative $x$ for most exponents — $(-4)^{1/2}$ is not a real number — and $0^{r}$ is undefined when $r\le 0.$ For $x>0,$ on the other hand, $x^{r}$ makes sense for every real $r,$ including irrational ones such as $x^{\pi}.$
Every power function $y=x^{r}$ passes through $(1,1),$ since $1^{r}=1,$ and its shape on $x>0$ depends only on $r{:}$ for $r>1$ it increases and bends upwards $(y=x^2);$ for $r=1$ it is the line $y=x;$ for $0<r<1$ it increases and bends downwards $(y=\sqrt{x});$ for $r=0$ it is the constant $y=1;$ and for $r<0$ it decreases, with $y\to\infty$ as $x\to 0$ and $y\to0$ as $x\to\infty$ $(y=1/x).$
Power functions are the standard building block for production functions in economics. A firm using $K$ units of capital and $L$ units of labour is commonly modelled by the Cobb–Douglas function $Y = A\,K^{\alpha}L^{\beta}$ with $A,\alpha,\beta>0{:}$ fixing $L,$ output is a power function of capital alone. Typically $0<\alpha<1,$ which by Fact 2.2 means the graph bends downwards — each extra machine adds less than the one before. That is diminishing returns, built into the model by the single assumption $\alpha<1.$
Doubling the input (an addition to the notes)
This example is not in the lecture notes; it is the lecture's own illustration of Fact 2.2.
A workshop with $L$ workers produces $Y = 30\,L^{1/2}$ units a day.
(a) By what factor does output grow when the workforce is doubled? When it is quadrupled?
(b) How much does one more worker add at $L = 100,$ and at $L = 400?$
Try it yourself first. Then open one step at a time.
(a) Compare the output at $2L$ with the output at $L{:}$ $$ \frac{30\,(2L)^{1/2}}{30\,L^{1/2}} = 2^{1/2} = \sqrt2 \approx 1.41\,. $$ The $L$ has cancelled: doubling the workforce multiplies output by $\sqrt2$ wherever you start. In the same way quadrupling it multiplies output by $4^{1/2} = 2.$ Four times the workforce, twice the output.
Nothing here used the exponent $\tfrac12$ in particular. For any power function $f(x) = Ax^{r},$ $$ \frac{f(2x)}{f(x)} = \frac{A\,(2x)^{r}}{A\,x^{r}} = 2^{r}\,. $$ That is what stays constant for a power function: the same proportional change in $x$ always gives the same proportional change in $y.$
(b) At $L=100$ output is $30\cdot 10 = 300,$ and with $101$ workers it is $30\sqrt{101}\approx 301.50{:}$ the extra worker adds about $1.50$ units. At $L=400$ output is $600,$ and with $401$ workers $30\sqrt{401}\approx 600.75{:}$ the extra worker adds about $0.75$ units.
So the graph rises all the time, but each step adds less than the one before. That is “increases and bends downwards” in Fact 2.2, with $r=\tfrac12$ between $0$ and $1{:}$ diminishing returns in numbers.
(a) By the factor $\sqrt2\approx1.41$ when doubled and by the factor $2$ when quadrupled, whatever $L$ is. (b) About $1.50$ units at $L=100$ and about $0.75$ units at $L=400.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's output as a function of its capital is $Y = 5K^{0.3}$ for $K > 0.$ What does Fact 2.2 say about the graph?
- Why is that answer tempting? Takes every positive exponent to behave like $x^{2}.$ Bending upwards needs $r > 1;$ here $0 < r < 1.$
- Why is that answer tempting? Reads $0.3 < 1$ as decay. That is the rule for the base of an exponential function, which comes in the next section. A power function decreases only when the exponent is negative.
- Why is that answer tempting? Reads the exponent as a coefficient: that would be $Y = 0.3K.$ The only power function that is a straight line has $r = 1.$
The exponent is $r = 0.3,$ between $0$ and $1,$ so by Fact 2.2 the graph increases and bends downwards. The factor $5$ stretches the graph vertically and does not change its shape.
In numbers: $K = 1, 2, 3$ give $Y = 5,$ about $6.16$ and about $6.95.$ The second unit of capital adds about $1.16,$ the third only about $0.80.$
Exponential functions
“Rises by 5 million a year” and “rises by 5% a year”: is that the same kind of growth?
In a power function the base varies and the exponent is fixed. Now we swap the two roles.
Guess first. The two families differ only in which letter moves. Before reading on, decide how a power function with a very large exponent compares with an exponential function whose base is barely above $1.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Compare $f(x) = x^{100}$ with $g(x) = 1.01^{\,x}$ for $x > 0$.
Which is bigger?
- Why is that answer tempting? Right winner, wrong quantifier. Misses that the two curves actually cross — and at x = 2, x¹⁰⁰ is already astronomically ahead.
- Why is that answer tempting? Compares the constants 100 and 1.01 directly. The intuition the question exists to break.
- Why is that answer tempting? Assumes two smooth increasing curves can keep swapping places. They cross exactly twice here — take logs and it becomes a line against a curve.
The argument uses the logarithm $\ln,$ which is the subject of the next section; come back to it afterwards if $\ln$ is new to you.
Take logs — legitimate, since $\ln$ is increasing: $$100 \ln x \quad \text{against} \quad x \ln 1.01$$
A logarithm against a straight line — and a line of any positive slope eventually beats any logarithm.
So $g$ wins, but not until $x \approx 117\,000.$
The point is not that exponentials are big, but that eventually and for a while are different claims.
Power against exponential
The blue curve is the power function $y = x^{r},$ the pink one the exponential function $y = a^{x}.$ The axes never move: $x$ runs from $0$ to $10.$
- Drag $r$ alone: $r = 1$ is a straight line, $r = 2$ a parabola. Only the blue curve moves.
- Drag $a$ alone: only the pink curve moves.
- Press Reset $(r = 3,$ $a = 2).$ The power function is ahead almost all the way. Find the two places where the curves cross: near $x = 1.4$ and near $x = 9.9.$
- Press Log axis. The exponential function becomes a straight line and the power function a curve that keeps flattening. Which of the two must win in the end?
The defining property is the exact analogue of Fact 1.3 for linear functions, with “adds” replaced by “multiplies”.
If $f(t)=Aa^{t},$ then for every $t,$ $$ f(t+1) = A a^{t+1} = a\cdot Aa^{t} = a\,f(t)\,. $$ So increasing $t$ by one unit multiplies the value by $a,$ no matter where you start, and $A=f(0).$ Hence $a>1$ gives growth, $0<a<1$ decay and $a=1$ a constant; writing $a = 1+p/100$ with $p>0,$ the function grows by $p\%$ per unit, and $a=1-p/100$ with $0<p<100$ makes it fall by $p\%$ per unit.
This is the whole difference between the two models of growth. Linear growth adds a fixed amount per period; exponential growth multiplies by a fixed factor. For populations, interest and prices the second is almost always the right starting assumption. A standard example fits Europe's population, in millions and with $t$ counted from 1960, by the linear model $P=5.1t+606;$ run backwards, that model reaches a negative population before 1841, whereas the constant proportional rate of $0.45\%$ per year gives $P(t)=606\cdot1.0045^{\,t},$ which does not.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A savings account holds $K_0$ today and grows by 4% each year.
Which formula gives the balance after $t$ years?
- Why is that answer tempting? Models a percentage rate as a constant absolute increase — the exact error Fact 2.5 is there to prevent. Also loses K₀ from the growth entirely.
- Why is that answer tempting? Uses the rate where the factor belongs. Gives a balance that collapses to nothing, which a five-second sanity check would catch.
- Why is that answer tempting? Power function instead of exponential — the two letters swapped. Predicts a balance of zero at t = 0.
Each year the balance is multiplied by $1.04.$ After $t$ years it has been multiplied $t$ times: $$K(t) = K_0 \cdot 1.04^{\,t}$$
B is the linear model. It is not merely less accurate — it says something different. On B the interest earned never changes; on A it grows every year, because you earn interest on the interest.
Two checks that cost nothing:
| value at $t = 0$ | after 1 year | |
|---|---|---|
| A | $K_0$ ✓ | $1.04K_0$ ✓ |
| B | $K_0$ ✓ | $K_0 + 0.04$ ✗ |
| C | $K_0$ ✓ | $0.04K_0$ ✗ |
| D | $0$ ✗ | — |
Put $t = 0$ and $t = 1$ into any growth formula before you trust it.
Compound interest
A deposit of $\text{\textdollar}K$ earning $p\%$ interest per year grows to $$ K\Bigl(1+\frac{p}{100}\Bigr)^{t} $$ after $t$ years.
Try it yourself first. Then open one step at a time.
At $8\%,$ one dollar becomes $1.08^{t}{:}$ after $10$ years about $\text{\textdollar}2.16,$ after $30$ years about $\text{\textdollar}10.06,$ and after $200$ years about $\text{\textdollar}4.8$ million.
Even a base only slightly above $1$ eventually produces enormous numbers — the single most counter-intuitive feature of exponential growth.
Decay is the same formula with $a<1{:}$ a car worth $P_0$ losing $20\%$ of its current value each year is worth $A(t)=P_0(0.8)^{t},$ about $0.33\,P_0$ after five years, and never worthless.
Growth by $p\%$ per year is multiplication by $1+p/100$ each year, so by $(1+p/100)^{t}$ after $t$ years: $1.08^{30}\approx 10.06.$ A fall by $20\%$ per year is multiplication by $0.8{:}$ $0.8^{5}\approx 0.33.$
Every base $a>0$ gives a different exponential function, and at first sight none is more natural than another. Once we have calculus, though, one base will make every formula simpler: the irrational number $e = 2.718281828459045\ldots,$ a little larger than $2.7.$
The notation $\exp(u)$ is used when $u$ is a complicated expression: compare $\exp\bigl(x^3+x\sqrt{x}-1/x+5\bigr)$ with the same thing as a superscript. You cannot avoid $e$ for long in statistics either — the normal density $\varphi(x)=\tfrac{1}{\sqrt{2\pi}}e^{-x^{2}/2},$ the bell curve, is the most important single function in the subject.
Nothing so far singles $e$ out: every base obeys these rules. What makes this particular base natural is a property of tangent lines, and it must wait until the derivative of $a^{x}$ is computed in Lecture 4.
Logarithms
How long until $1000 at $8\%$ interest has become $10 000?
Economists constantly ask questions of the form: how long until…? How long until the price level triples? How long until $1000 at $8\%$ becomes $10 000? Each of these asks us to solve $a^{x} = b$ for $x$ — that is, to undo an exponential. The logarithm is the tool that does it.
Memorise the sentence in italics; every computation below is an unpacking of it. So $\ln 1=0$ because $e^{0}=1;$ $\ln e=1;$ $\ln(1/e)=-1;$ and $\ln 4\approx1.386$ lies between $1$ and $2,$ as one sees without a calculator from $e^{1}\approx2.7<4<7.4\approx e^{2}.$ Note that $\ln a$ is defined only for $a>0,$ because $e^{x}$ is always positive.
For all $x,y>0$ and all real $p,$ $$ \begin{aligned} \ln(xy) &= \ln x + \ln y\,,\\ \ln\frac{x}{y} &= \ln x - \ln y\,,\\ \ln x^{p} &= p\,\ln x\,, \end{aligned} $$ and $\ln$ and $\exp$ undo each other: $\ln e^{x}=x$ for all real $x,$ and $e^{\ln x}=x$ for all $x>0.$
The first rule. For the first rule, use (2.1) three times together with $e^{s}e^{t}=e^{s+t}{:}$ $$ e^{\ln(xy)} = xy = e^{\ln x}\,e^{\ln y} = e^{\ln x+\ln y}\,. $$ Since $e^{u}=e^{v}$ forces $u=v,$ we get $\ln(xy)=\ln x+\ln y.$
The other two. The other two rules follow in the same way from $e^{s}/e^{t}=e^{s-t}$ and $(e^{s})^{t}=e^{st}.$
The third rule is the one that matters: it brings the exponent down, turning an equation in which the unknown sits upstairs into an ordinary linear equation.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Which of these equals $\ln 12?$
- Why is that answer tempting? Splits $12 = 10 + 2$ and takes the logarithm of each part. But $\ln 10 + \ln 2 = \ln(10\cdot 2) = \ln 20{:}$ there is no rule for the logarithm of a sum (Watch out 2.10).
- Why is that answer tempting? Keeps the product as a product. A product inside the logarithm becomes a sum outside: $\ln 3\cdot\ln 4 \approx 1.52,$ while $\ln 12 \approx 2.48.$
- Why is that answer tempting? Reads $\ln 2^{2}$ as $(\ln 2)^{2}.$ The exponent comes down as a factor, $\ln 2^{2} = 2\ln 2;$ it does not move outside the logarithm as a power.
Write $12 = 2^{2}\cdot 3$ and use two rules of Fact 2.9, one at a time: $$ \ln 12 = \ln 2^{2} + \ln 3 = 2\ln 2 + \ln 3\,. $$ A calculator confirms it: $2\cdot 0.693 + 1.099 \approx 2.485 \approx \ln 12.$ The other three give $\ln 20\approx 3.00,$ about $1.52$ and about $1.58.$
Try the method once yourself before the worked example. The equation is the one this section opened with.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Solve for $x$: $$1000 \cdot (1.08)^{\,x} = 10\,000$$
- Why is that answer tempting? Treats the exponent as a factor. Usually means they never got as far as taking a logarithm.
- Why is that answer tempting? Collapses (ln 10)/(ln 1.08) into ln(10/1.08) — a quotient of logarithms read as the logarithm of a quotient. The most common log error after ln(x+y).
- Why is that answer tempting? Took the log on one side only. Half the method, applied asymmetrically.
C is the one to look at. These are different numbers: $$\frac{\ln 10}{\ln 1.08} \approx 29.9 \qquad \ln\!\left(\frac{10}{1.08}\right) \approx 2.2$$
$\dfrac{\ln x}{\ln y}$ is a ratio of two logarithms. $\ln\dfrac{x}{y}$ is the logarithm of a ratio. The rule $\ln(x/y) = \ln x - \ln y$ turns the second into a difference — it says nothing whatever about the first.
The interpretation: at 8% a year, money takes just under 30 years to grow tenfold.
Two equations with the unknown in the exponent
Solve for $x{:}$ (a) $(1.08)^{x}=10;$ (b) $e^{x}+4e^{-x}=4.$
Try it yourself first. Then open one step at a time.
(a) Taking $\ln$ of both sides gives $x\ln 1.08 = \ln 10,$ so $$ x = \frac{\ln 10}{\ln 1.08}\approx 29.9\,. $$ So it takes just under $30$ years for a deposit to grow tenfold at $8\%$ interest — which answers the second question we opened this subsection with.
(b) Here taking $\ln$ leads nowhere, because $\ln(e^{x}+4e^{-x})$ cannot be simplified (Warning 2.10). Instead put $u=e^{x},$ so that $e^{-x}=1/u{:}$ $$ \begin{aligned} &u+\frac{4}{u}=4\\ &\Longrightarrow\quad u^{2}-4u+4=0\\ &\Longrightarrow\quad (u-2)^{2}=0\,. \end{aligned} $$
So $u=2,$ that is $e^{x}=2,$ and $x=\ln 2.$
(a) $x = \dfrac{\ln 10}{\ln 1.08}\approx 29.9.$
(b) $x = \ln 2\approx 0.69.$ Check: $e^{\ln 2} + 4e^{-\ln 2} = 2 + 4\cdot\tfrac12 = 4.$
The equation $u=\ln x$ says exactly the same thing as $x=e^{u},$ so the graph of $y=\ln x$ is the graph of $y=e^{x}$ reflected in the line $y=x{:}$ defined only for $x>0,$ strictly increasing, crossing the $x$-axis at $x=1,$ and falling steeply towards $-\infty$ as $x$ approaches $0.$ Occasionally a base other than $e$ is convenient. For $a>0,$ $a\ne1,$ write $\log_a x$ for the power to which $a$ must be raised to give $x,$ so that $a^{\log_a x}=x;$ taking $\ln$ of this identity gives $\log_a x\cdot\ln a=\ln x,$ that is $\log_a x = \ln x/\ln a.$ So all logarithms are proportional to one another, and $\log_a$ obeys exactly the same three rules as ${\ln}.$
Let $a>1$ and $A>0.$ The exponential function $f(t)=Aa^{t}$ has a unique doubling time $t^{*},$ the solution of $a^{t^{*}}=2,$ namely $$ t^{*} = \frac{\ln 2}{\ln a}\,. $$ It does not depend on $A,$ nor on the moment from which you start measuring.
We want $f(t+t^{*})=2f(t),$ that is $Aa^{t}a^{t^{*}}=2Aa^{t}.$ Cancelling $Aa^{t},$ which is never zero, leaves $a^{t^{*}}=2$ — an equation in which neither $A$ nor $t$ appears. Taking $\ln$ gives $t^{*}\ln a=\ln 2.$
Doubling times, and the rule of 70
How long does a population take to double at the growth rates of Zimbabwe and Kenya in the 1970s and 1980s?
Try it yourself first. Then open one step at a time.
Zimbabwe's population grew at about $3.5\%$ per year in the 1970s and 1980s, so $a=1.035$ and $$ t^{*} = \frac{\ln 2}{\ln 1.035}\approx 20.1 \text{ years.} $$
Kenya's growth rate in the 1980s was $4.2\%,$ giving $t^{*}=\ln2/\ln1.042\approx 16.9$ years.
(Correction by the app: $\ln 2/\ln 1.042 = 16.85$ to two decimals, so to one decimal the doubling time is $16.8$ years, not $16.9.)$
A useful mental shortcut: for small growth rates $p\%,$ $\ln(1+p/100)\approx p/100,$ and $\ln 2 \approx 0.69,$ so $$ t^{*}\approx \frac{70}{p}\,. $$ This is the “rule of $70$” that appears in every introductory economics course. At $3.5\%$ it predicts $20$ years; at $4.2\%$ it predicts $16.7.$ Both are close.
About $20.1$ years at $3.5\%$ and about $16.8$ years at $4.2\%.$ The rule of $70$ gives $70/3.5 = 20$ and $70/4.2\approx 16.7.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A quantity grows exponentially and doubles every $12$ years. How long does it take to become $8$ times as large?
- Why is that answer tempting? Counts $8/2 = 4$ doublings. But four doublings multiply by $2^{4} = 16;$ the number of doublings is the exponent in $2^{k} = 8,$ not the quotient $8/2.$
- Why is that answer tempting? Reasons proportionally: twice as large takes $12$ years, so $8$ times as large takes $8\cdot 12.$ That is linear growth. Exponential growth multiplies by $2$ in every period of $12$ years.
- Why is that answer tempting? The starting value $A$ cancels, as in the reason for Fact 2.13: the time needed to grow by a given factor is the same wherever you start.
$8 = 2^{3},$ so the quantity must double three times, and by Fact 2.13 each doubling takes $12$ years wherever it starts: $3\cdot 12 = 36$ years.
With the formula: the base satisfies $a^{12}=2,$ and $a^{t}=8$ gives $t = \ln 8/\ln a = 3\ln 2/\ln a = 3\cdot 12.$
The derivative
A straight line has one slope. What is the slope of a curve at a single point?
We now change subject. For a straight line $y=px+q$ we have a perfectly good measure of steepness: the slope $p,$ which is the same everywhere. For a curve, steepness clearly varies from point to point — so what number should we attach to a single point? The answer is to look at the straight line that best matches the curve there.
Definition 2.15 is already enough to read a derivative off a picture: imagine the tangent at each point and compare the slopes.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
The graph shows $y = f(x)$, with four points marked.
At which of them is $f'(x)$ largest?
- Why is that answer tempting? Picks a point where the curve is clearly rising, without comparing it with d. Close but beaten — f′(a) = +2.4 against +3.6.
- Why is that answer tempting? Reads off where f is largest instead of where f′ is largest. The single most expensive confusion in the course — and at b the tangent is horizontal, so f′(b) = 0.
- Why is that answer tempting? Reads 'largest' as 'steepest' and ignores the sign. The curve is falling at c, so f′(c) is negative — the smallest of the four, not the largest.
Lay a ruler along the curve at each point. The slopes of the four tangents are about $+2.4$ at $a,$ $0$ at $b,$ $-3.1$ at $c$ and $+3.6$ at $d.$ (The same curve is drawn with its four tangents at the start of the last section.)
$f$ and $f'$ are different functions. $b$ is where $f$ is largest, and it is exactly where $f'$ is zero. High and steep are unrelated.
Slopes are signed. $c$ is the steepest-looking point on the picture, and its derivative is the smallest of the four.
Definition 2.15 is geometric and vague: “just touches” is not something we can compute with. Here is the idea that makes it precise. Fix $P=(a,f(a))$ on the graph and take a second point $Q$ on the graph, a little to the right, with $x$-coordinate $a+h.$ The line through $P$ and $Q$ is called a secant, and we know its slope exactly, by the formula (1.1) from Lecture 1: $$ \frac{f(a+h)-f(a)}{h}\,. \tag{2.2} $$ This expression is called the Newton quotient of $f$ at $a.$ Now slide $Q$ along the curve towards $P,$ that is, let $h$ shrink towards $0.$ The secant pivots about $P$ and settles down onto the tangent, so the slope of the tangent should be the value that (2.2) approaches.
Note carefully that we cannot simply substitute $h=0$ into (2.2): both numerator and denominator would be $0,$ and $0/0$ is not a number. The whole content of the definition is that the quotient can nevertheless settle on a definite value as $h$ gets small. The word “limit” is made precise in Lecture 5; for now, intuition suffices.
Let h → 0
Two numbers are shown: the slope of the secant and the derivative.
- Press h → 0 and watch the slope of the secant close in on the derivative.
- Look at where $h$ stops: at a small positive number, not at $0.$ At $h = 0$ there would be nothing to compute.
- Move $a$ to the left of the origin and press h → 0 again. The derivative is negative there, and the secant follows.
- Press the button f(x) = x² once: the function changes to $x^{3}.$ Put $a = 0.$ The tangent is horizontal, although the curve rises on both sides. We return to this in the last section.
The derivative of $x^{2}$ from the definition
Compute $f'(a)$ for $f(x)=x^{2}.$
Try it yourself first. Then open one step at a time.
Following the recipe: $$ \begin{aligned} f(a+h) &= (a+h)^{2}\\ &= a^{2}+2ah+h^{2}\,, \end{aligned} $$
The difference: $$ \begin{aligned} f(a+h)-f(a) &= 2ah+h^{2}\\ &= h(2a+h)\,, \end{aligned} $$
Divide by $h$ and cancel: $$ \begin{aligned} \frac{f(a+h)-f(a)}{h} &= \frac{h(2a+h)}{h}\\ &= 2a+h\,. \end{aligned} $$
As $h\to0,$ this tends to $2a.$ Hence $f(x)=x^{2}$ has $f'(x)=2x.$
Sanity check against the picture: $f'(x)<0$ for $x<0,$ $f'(0)=0$ and $f'(x)>0$ for $x>0$ — exactly the behaviour of the parabola, which falls, flattens at its vertex, then rises.
$f'(a) = 2a$ for every $a;$ as a function, $f'(x) = 2x.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Let $f(x) = x^{2} + 3x$ and $a = 1.$ What does the Newton quotient look like after step 4 of Method 2.18, simplified and with $h \ne 0?$
- Why is that answer tempting? That is $f'(1),$ the value after step 5. The Newton quotient is the slope of a secant and still contains $h;$ only the limit removes it.
- Why is that answer tempting? Puts $h = 0$ into the quotient before cancelling. The quotient is not defined at $h = 0,$ which is exactly why $h$ is cancelled first and the limit taken afterwards.
- Why is that answer tempting? Loses the term $3x{:}$ this is the Newton quotient of $x^{2}$ alone at $a = 1.$ The term $3x$ adds $3h$ to the difference and so $3$ to the quotient.
Follow the Method. $f(1+h) = (1+h)^{2} + 3(1+h) = 4 + 5h + h^{2}$ and $f(1) = 4,$ so the difference is $5h + h^{2} = h(5 + h){:}$ everything without $h$ has cancelled. Dividing by $h \ne 0$ leaves $5 + h.$
Only now let $h \to 0{:}$ $f'(1) = 5.$
Quadratics, the tangent line and rates of change
Lecture 1 found the vertex of a parabola by completing the square. Can the derivative find it too?
Carrying out this procedure every time would be intolerable, and Lecture 3 develops rules that make it unnecessary. One case is worth recording now, because it covers everything in Lecture 1.
The Newton quotient is $$ \begin{aligned} &\frac{a(x+h)^2+b(x+h)+c-ax^2-bx-c}{h}\\ &\quad= \frac{2axh+ah^{2}+bh}{h} = 2ax+b+ah\,, \end{aligned} $$ which tends to $2ax+b$ as $h\to0.$
This closes a circle. In Lecture 1 we found by completing the square that a quadratic has its vertex at $x^{*}=-b/2a.$ Fact 2.20 says that $f'(x)=0$ exactly when $2ax+b=0,$ that is when $x=-b/2a$ — the same point. The vertex is where the tangent is horizontal. This is the first glimpse of the central technique of the course: to optimise, differentiate and set the derivative to zero.
It also gives the tangent itself. Applying the point–slope formula (1.2) at the point $(a,f(a))$ with slope $f'(a),$ the tangent to $y=f(x)$ at that point is the line $y = f(a) + f'(a)(x-a).$
A tangent line, step by step (an addition to the notes)
This example is not in the lecture notes; it is the one the lecture itself works through.
Find the tangent to $y = x^{2}$ at $x = \tfrac12.$
Try it yourself first. Then open one step at a time.
The point. $f\bigl(\tfrac12\bigr) = \tfrac14,$ so the tangent passes through $\bigl(\tfrac12, \tfrac14\bigr).$
The slope. By Fact 2.20, $f'(x) = 2x,$ so $f'\bigl(\tfrac12\bigr) = 1.$ Differentiate first, put in the number second.
The line. Put the two numbers into $y = f(a) + f'(a)(x-a){:}$ $$ y = \tfrac14 + 1\cdot\bigl(x - \tfrac12\bigr) = x - \tfrac14\,. $$
Check. At $x = \tfrac12$ the line gives $\tfrac12 - \tfrac14 = \tfrac14,$ the value of $f{:}$ it passes through the point. And its slope is $1.$
Read the formula as an instruction: start at the point on the curve, then walk along at the curve's own slope. Near $x = \tfrac12$ the line imitates $f$ well: at $x = 0.6$ the curve has height $0.36$ and the line $0.35.$
$y = \tfrac14 + \bigl(x - \tfrac12\bigr),$ that is $y = x - \tfrac14.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Which line is the tangent to $f(x) = x^{2} - 4x$ at $x = 3?$
- Why is that answer tempting? Uses $f(3) = -3$ as the intercept $m$ in $y = kx + m.$ This line passes through $(3, 3),$ not through $(3, -3).$
- Why is that answer tempting? The two numbers have changed places: $f'(3) = 2$ is used as the height and $f(3) = -3$ as the slope.
- Why is that answer tempting? Leaves the derivative as the function $2x - 4$ and does not put in $x = 3.$ The result contains $x^{2},$ so it is not a line at all.
Two numbers are needed. The point: $f(3) = 9 - 12 = -3.$ The slope: by Fact 2.20, $f'(x) = 2x - 4,$ so $f'(3) = 2.$ Hence $$ y = -3 + 2(x - 3)\,, $$ which is $y = 2x - 9.$ Check: at $x = 3$ it gives $-3.$
The derivative is written in several ways, all meaning the same thing: $$ f'(x)\,,\qquad y'\,,\qquad \frac{dy}{dx}\,,\qquad \frac{d}{dx}f(x)\,, $$ the last two being Leibniz's differential notation; read $\dfrac{d}{dx}$ as the instruction “differentiate with respect to $x$”. When the variable is time a dot is common, $\dot{x}(t),$ and letters change with the context: $P(t)=t^{2}$ gives $P'(t)=2t.$
The Newton quotient is the average rate of change of $f$ between $a$ and $a+h,$ so its limit $f'(a)$ is the instantaneous rate of change at $a;$ and for small $h,$ $f(a+h)-f(a)\approx f'(a)\cdot h.$
In economics this is exactly what the word marginal means. If $C(Q)$ is total cost, $C'(Q)$ is marginal cost, the extra cost of producing roughly one more unit; if $R(Q)$ is revenue, $R'(Q)$ is marginal revenue. So a firm with $C(x)=p+qx^{2}$ has marginal cost $C'(x)=2qx$ by Fact 2.20: zero at $x=0$ and increasing with output. The fixed cost $p$ has vanished, as it must — fixed costs cannot affect the cost of one more unit.
Recap The economics words, once and for good
If the words stop you and not the mathematics, here they are. They stay the same for the whole course.
| word | symbol | what it means |
|---|---|---|
| quantity, demand | $Q$ | units sold |
| price | $P$ | price per unit |
| revenue | $R = P\cdot Q$ | money in |
| cost | $C(Q)$ | money out |
| profit | $\pi = R - C$ | what is left |
| marginal anything | $R'(Q),$ $C'(Q)$ | the effect of one more unit |
There is no new mathematics in the table. “Marginal” is today's word: it means derivative.
Increasing and decreasing functions
At the top of a hill $f$ is as large as it gets. What is $f'$ there?
Look at the signs in the picture. Where the curve rises, at $a$ and at $d,$ the derivative is positive. Where it falls, at $c,$ the derivative is negative. At the top, at $b,$ the tangent is horizontal and the derivative is $0.$ This section turns that observation into a method. First the words have to be exact.
Note that an “increasing” function is allowed to have flat stretches — which is why such functions are sometimes called nondecreasing. Checking these definitions directly, by comparing $f(x_1)$ and $f(x_2)$ for all pairs of points, is hopeless in practice. The derivative gives a test that is genuinely usable.
Let $f$ be differentiable on an interval $I.$ Then $$ \begin{aligned} &f'(x)\ge 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is increasing on } I\,,\\ &f'(x)\le 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is decreasing on } I\,,\\ &f'(x)= 0 \text{ for all } x\in I\\ &\qquad\iff f \text{ is constant on } I\,. \end{aligned} $$ Moreover $f'(x)>0$ for all $x\in I$ implies that $f$ is strictly increasing on $I,$ and similarly $f'(x)<0$ implies strictly decreasing.
Given that $f'(x)$ is the slope of the tangent, the equivalences are close to obvious: a graph whose tangents all point upwards can only rise. (A careful proof needs the mean value theorem, which this course does not cover.)
Guess first. The first three lines of Fact 2.23 run in both directions. The last sentence is stated in one direction only. Is that an accident of the wording?
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A student writes:
"$f$ is strictly increasing on $I$, so $f'(x) > 0$ at every point of $I$."
Is that right?
- Why is that answer tempting? Reads the last sentence of Fact 2.23 backwards: the implication runs from f′ > 0 to strictly increasing, not the other way. The commonest misuse of the whole section.
- Why is that answer tempting? Hedged version of A. x³ is a polynomial, so the hedge does not even rescue it.
- Why is that answer tempting? Overshoots. If f′(x₀) < 0 then f really does decrease near x₀ — negative is impossible, zero is not. Tempting for anyone who spots that A is wrong but not why.
$f(x) = x^3$ is strictly increasing everywhere, yet $f'(x) = 3x^2$ gives $f'(0) = 0$ — it flattens at the origin without ever turning round.
So the last sentence of Fact 2.23 runs one way only: $$ \begin{aligned} &f' > 0 \text{ on } I\\ &\quad\Longrightarrow\; f \text{ strictly increasing on } I \end{aligned} $$
With $\ge$ it runs both ways: $$f' \ge 0 \text{ on } I \iff f \text{ increasing on } I$$
Why D fails: a negative derivative makes $f$ genuinely decrease near that point. Zero slope is survivable; negative slope is not.
A sign diagram for a cubic
Where is $f(x)=-\tfrac13x^{3}+2x^{2}-3x+1$ increasing and decreasing? You may use that its derivative is $f'(x)=-x^{2}+4x-3.$
Try it yourself first. Then open one step at a time.
Solve $f'(x)=0{:}$ the roots of $-x^2+4x-3$ are $x=1$ and $x=3,$ so $$ \begin{aligned} f'(x) &= -(x-1)(x-3)\\ &= (x-1)(3-x)\,. \end{aligned} $$
Now read off the signs of the two factors on either side of $1$ and $3{:}$ $$ \begin{array}{l|ccccc} & x<1 & x=1 & 1<x<3 & x=3 & x>3\\\hline x-1 & - & 0 & + & + & +\\ 3-x & + & + & + & 0 & -\\\hline f'(x) & - & 0 & + & 0 & - \end{array} $$
Hence $f$ is decreasing on $(-\infty,1],$ increasing on $[1,3],$ and decreasing again on $[3,\infty).$
The points $x=1$ and $x=3,$ where $f'$ changes sign, are a local minimum and a local maximum respectively — the turning points of the cubic. A week ago we could only have found them by trial and error; now they drop out of a sign diagram.
$f$ is decreasing on $(-\infty,1],$ increasing on $[1,3]$ and decreasing on $[3,\infty).$ The endpoints belong to the intervals: Method 2.25 reports $f'\ge0$ and $f'\le0.$
What you should take away
- A power function $Ax^{r}$ has a fixed exponent and a varying base; an exponential function $Aa^{t}$ has a fixed base and a varying exponent. Do not confuse them.
- Linear growth adds a fixed amount per period; exponential growth multiplies by a fixed factor $a$ per period. If $a=1+p/100,$ the growth is $p\%$ per period.
- $\ln a$ is the power of $e$ needed to give $a{:}$ $e^{\ln a}=a.$ It obeys $\ln(xy)=\ln x+\ln y,$ $\ln(x/y)=\ln x-\ln y$ and $\ln x^{p}=p\ln x$ — but there is no rule for $\ln(x+y).$ The third rule is what lets you solve $a^{x}=b.$
- The doubling time of $Aa^{t}$ is $t^{*}=\ln2/\ln a,$ independent of $A$ and of the starting date; for small rates, $t^{*}\approx 70/p.$
- The derivative $f'(a)=\lim_{h\to0}\bigl(f(a+h)-f(a)\bigr)/h$ is the slope of the tangent at $(a,f(a)),$ and the instantaneous rate of change of $f$ there — in economics, the marginal quantity. The tangent is $y=f(a)+f'(a)(x-a).$
- $f(x)=ax^{2}+bx+c \Rightarrow f'(x)=2ax+b;$ in particular $f'(x)=0$ at $x=-b/2a,$ the vertex found by completing the square in Lecture 1.
- $f'\ge0$ on an interval means increasing, $f'\le0$ means decreasing. But $f'>0$ is only sufficient for strict increase, not necessary: recall $x^{3}$ at $x=0.$
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — which letter moves
Which of the following define exponential functions of $x?$ $y=3^{x};$ $y=x^{\sqrt2};$ $y=(\sqrt2)^{x};$ $y=x^{x};$ $y=(1/2)^{x}.$
One way to think about it
Three of them: $3^{x},$ $(\sqrt2)^{x}$ and $(1/2)^{x}.$ In each the base is a fixed positive number and the variable sits in the exponent, as Definition 2.3 requires; the last one has a base below $1,$ so it describes decay, but it is exponential all the same.
$y=x^{\sqrt2}$ is a power function (Definition 2.1): the base varies and the exponent is fixed. That the exponent is irrational changes nothing. $y=x^{x}$ is neither: both the base and the exponent vary.
Discussion question 2 — falling by 50% twice
A quantity falls by $50\%$ each year. After two years it has fallen in total by
- (A) $100\%;$
- (B) $75\%;$
- (C) $50\%;$
- (D) $25\%.$
One way to think about it
(B). A fall of $50\%$ is multiplication by $0.5$ (Fact 2.5), so after two years the factor is $0.5^{2}=0.25.$ A quarter is left, and the total fall is $75\%.$
(A) adds the percentages, which is the linear model: it would leave nothing after two years and a negative quantity after three. The second fall of $50\%$ is taken from what is left, not from the original amount. (D) is the share that remains, not the share that has gone.
Discussion question 3 — which rules are rules
Which of these is always true?
- (A) $(\ln A)^{4} = 4\ln A;$
- (B) $\ln B = 2\ln\sqrt{B};$
- (C) $\ln A^{10}-\ln A^{4} = 3\ln A^{2};$
- (D) $\ln\dfrac{A+B}{C} = \ln A+\ln B-\ln C.$
One way to think about it
Two of them, (B) and (C), for all positive $A$ and $B.$
(B): $\sqrt{B}=B^{1/2},$ so $2\ln\sqrt{B}=2\cdot\tfrac12\ln B=\ln B$ by the third rule of Fact 2.9.
(C): the left side is $10\ln A-4\ln A=6\ln A,$ and the right side is $3\cdot 2\ln A=6\ln A.$
(A) confuses $(\ln A)^{4}$ with $\ln A^{4}.$ For $A=e$ the left side is $1^{4}=1$ and the right side is $4.$
(D) is Watch out 2.10: there is no rule for $\ln(A+B).$ For $A=B=C=1$ the left side is $\ln 2\approx0.69$ and the right side is $0.$
Discussion question 4 — two doubling times
Two quantities grow exponentially with doubling times $10$ and $20$ years. After $60$ years, by what factor has each grown? Does the one that started smaller necessarily stay smaller?
One way to think about it
Sixty years is six doubling times for the first and three for the second, so the factors are $2^{6}=64$ and $2^{3}=8.$
No. If the smaller one is the one that doubles every $10$ years, it overtakes the other in the end, however far behind it starts. Take the values $1$ and $4$ at the start: the quantities are $2^{t/10}$ and $4\cdot2^{t/20}=2^{2+t/20},$ and they are equal when $t/10=2+t/20,$ that is after $40$ years (both are then $16).$ After that the first is larger for ever.
If instead the smaller one has the longer doubling time, it does stay smaller: the ratio between the two only grows.
Discussion question 5 — a horizontal tangent
If $f'(a)=0,$ then $f$ has a maximum or a minimum at $a.$
- (A) True.
- (B) False.
- (C) True only if $f$ is a polynomial.
- (D) True only if $f'$ changes sign at $a.$
One way to think about it
(B), false. The counterexample is the one from Watch out 2.24: $f(x)=x^{3}$ has $f'(0)=0,$ yet it is strictly increasing, so it has neither a maximum nor a minimum at $0.$ Since $x^{3}$ is a polynomial, (C) does not rescue the claim.
(D) points at what is missing. If $f'$ changes sign at $a,$ as it does at $x=1$ and $x=3$ in Example 2.26, then $f$ does turn round there. A horizontal tangent alone is not enough. Later lectures make this precise; for now the point is that $f'(a)=0$ tells you where to look, not what you will find.
Discussion question 6 — the sign of the marginal profit
A firm's profit satisfies $\pi'(Q)>0$ for $Q<40$ and $\pi'(Q)<0$ for $Q>40.$ What should the firm do, and why does it not need to know $\pi$ itself?
One way to think about it
Produce $Q=40.$ By Fact 2.23 the profit is strictly increasing for $Q<40$ and strictly decreasing for $Q>40,$ so every other quantity gives a smaller profit than $40$ does.
Only the sign of $\pi'$ was used. The firm needs to know whether one more unit adds to the profit or takes away from it — whether the marginal profit is positive or negative — and not how large the profit is. In particular a fixed cost, which shifts $\pi$ up or down without changing $\pi',$ has no influence on the best quantity.