A sign change forces a root
A function has $f(-1)=-1$ and $f(1)=1.$ Must it be zero somewhere in between?
Draw a curve from a point below the $x$-axis to a point above it without lifting the pen. Somewhere it has to cross the axis. The first theorem of this lecture, the intermediate value theorem, is that observation made precise — and the precise version says exactly what “without lifting the pen” has to mean.
Let $f$ be continuous on the closed interval $[a,b]$. Then $f$ takes every value between $f(a)$ and $f(b)$ somewhere on $[a,b]$. In particular, if $f(a)$ and $f(b)$ have opposite signs, there is at least one $c$ in $(a,b)$ with $f(c)=0$.
This is how you learn that an equation has a solution before having any idea how to solve it — which is most of the time, since almost no equation met in practice can be solved in closed form.
Guess first. The next question is the one under the heading of this section. Before you answer, try to find a function that satisfies $f(-1)=-1$ and $f(1)=1$ and is never zero in between.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A function $f$ satisfies $f(-1) = -1$ and $f(1) = 1$.
Must there be a $c$ in $(-1, 1)$ with $f(c) = 0$?
- Why is that answer tempting? Drops the hypothesis. A sign change on its own guarantees nothing at all, and there is a one-line counterexample from Lecture 5.
- Why is that answer tempting? Asks for more than is needed, and still not quite the right thing: differentiability on the open interval says nothing about what happens at the endpoints. The instinct to ask for a condition is right; the condition is continuity on the closed interval.
- Why is that answer tempting? Over-corrects. With continuity the guarantee is absolute; the point is that the hypothesis is doing real work, not that the theorem is weak.
Take $f(x) = 1/x$: it runs from $-1$ to $1$ and is never zero.
The theorem is not violated — $1/x$ is not continuous on $[-1,1]$, since it is not even defined at $0$. Continuity is the whole hypothesis.
On C: differentiability would be enough (it implies continuity), but it asks far more than needed — and on the open interval it says nothing about the endpoints.
A root nobody has to compute, and a counterexample
(a) Show that $x^{6}+3x^{2}-2x-1=0$ has a solution between $0$ and $1.$
(b) A function satisfies $f(-1)=-1$ and $f(1)=1.$ Must it be zero somewhere in between?
Try it yourself first. Then open one step at a time.
(a) Put $f(x)=x^{6}+3x^{2}-2x-1.$ It is a polynomial, hence continuous, and $f(0)=-1$ while $f(1)=1+3-2-1=1.$ The signs differ, so by Fact 6.1 there is a root in $(0,1)$ — and nobody had to solve a sextic.
(b) No. Take $f(x)=1/x,$ which runs from $-1$ at $x=-1$ to $1$ at $x=1$ and is never zero. The theorem is not violated, because $1/x$ is not continuous on $[-1,1]{:}$ it is not even defined at $0.$ Continuity is the entire hypothesis, and this is the same function that failed to have a limit at $0$ in Example 5.11.
(a) There is a root in $(0,1),$ because $f(0)=-1<0<1=f(1)$ and $f$ is continuous.
(b) No: $1/x$ changes sign on $[-1,1]$ and is never zero. It is not continuous there, so Fact 6.1 does not apply.
When a hypothesis in a theorem looks like decoration, the useful reflex is to go looking for the counterexample that put it there.
Fact 6.1 gives at least one root. Often what is wanted is exactly one, and that is a second and separate job: monotonicity supplies it, since a strictly increasing function takes each value at most once. The pairing is worth recognising as a pattern, because an examiner who writes “show that the equation has exactly one solution” has set two tasks and not one.
Exactly one solution: two tasks
Show that $2x = 5e^{-x}(1+x^{2})$ has exactly one solution.
Try it yourself first. Then open one step at a time.
Put $g(x) = 2x - 5e^{-x}(1+x^{2}),$ so that solutions are the zeros of $g.$
At least one. $g(0) = -5 < 0$ and $g(2) = 4 - 25e^{-2} \approx 0.62 > 0,$ and $g$ is continuous, so Fact 6.1 gives a zero in $(0,2).$
At most one. Differentiating with the product and chain rules, $$ \begin{aligned} g'(x) &= 2 + 5e^{-x}\bigl(x^{2}-2x+1\bigr)\\ &= 2 + 5e^{-x}(x-1)^{2} > 0 \end{aligned} $$ for every $x,$ so $g$ is strictly increasing by Fact 2.23 and cannot take the value $0$ twice.
Exactly one solution, therefore, somewhere in $(0,2)$ — and no formula for it anywhere.
Exactly one solution, and it lies in $(0,2).$ The sign change gives at least one; $g'>0$ gives at most one.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
How many real solutions does the equation $$ x^{3}+x-1=0 $$ have?
- Why is that answer tempting? Stops after the first of the two tasks. The derivative settles the second without solving anything: $3x^{2}+1>0,$ as in Example 6.4.
- Why is that answer tempting? A polynomial of degree $3$ has at most three roots (Lecture 1, after Fact 1.17). How many it really has is a separate question.
- Why is that answer tempting? The right number for the wrong reason. Fact 6.1 promises at least one and says nothing about how many (Watch out 6.2).
Newton's method
We know that $x^{2}=2$ has a solution between $1$ and $2.$ How do we get its digits?
The intermediate value theorem says a root is in there and gives no hint where. Newton's method goes and gets it, and it is nothing but the linear approximation of Lecture 4 used as an algorithm.
Start from a guess $x_{0}.$ Near $x_{0}$ the function is well approximated by its tangent, $y = f(x_{0}) + f'(x_{0})(x-x_{0}).$ We cannot solve $f(x)=0,$ but we can certainly solve the tangent equal to zero: setting $y=0$ and rearranging gives $x = x_{0} - f(x_{0})/f'(x_{0}).$ Call that $x_{1},$ and repeat.
Guess first. One step by hand, before the worked example does it.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Apply one step of Newton's method to $$f(x) = x^{2} - 2, \qquad x_{0} = 2$$
What is $x_{1}$?
- Why is that answer tempting? Added instead of subtracting. A sketch settles it: the root is below $2,$ so the step must go down.
- Why is that answer tempting? Divided by $x_{0}$ rather than by $f'(x_{0}){:}$ $2-2/2.$ Confuses the value with the slope.
- Why is that answer tempting? Gives the exact root. Newton lands on it in one step only if $f$ is linear; one step gets close, not there. Still, this answer shows that you know where the method is going.
Keep going and watch the digits double:
| $n$ | $x_n$ | error |
|---|---|---|
| 0 | $2$ | $0.59$ |
| 1 | $1.5$ | $0.086$ |
| 2 | $1.41667$ | $0.0025$ |
| 3 | $1.414216$ | $0.0000021$ |
Three steps from a crude guess, and $\sqrt2$ to six decimal places.
What $x_1 = 1.5$ actually is: the point where the tangent at $x=2$ crosses the axis. Every step is one linear approximation, solved.
A square root in three steps, and a two-cycle
(a) Find $\sqrt{2}$ by applying (6.1) to $f(x)=x^{2}-2$ from $x_{0}=2.$
(b) Apply it to $f(x)=x^{3}-2x+2$ from $x_{0}=0.$
Try it yourself first. Then open one step at a time.
(a) Here $f'(x)=2x,$ so $x_{n} = x_{n-1} - (x_{n-1}^{2}-2)/2x_{n-1}.$ From $x_{0}=2{:}$
iterate error $x_{0}=2$ 0.59 $x_{1}=1.5$ 0.086 $x_{2}=1.416667$ 0.0025 $x_{3}=1.4142157$ 0.0000021 Three steps from a crude guess give $\sqrt2$ to within a couple of millionths. Notice that the error does not shrink steadily, it collapses: the number of correct digits roughly doubles at each step, which is why the method has survived three centuries.
(b) Here $f'(x)=3x^{2}-2.$ From $x_{0}=0$ we get $x_{1} = 0 - 2/(-2) = 1,$ and then $x_{2} = 1 - 1/1 = 0.$ The iteration runs $0,1,0,1,\ldots$ for ever and never goes anywhere at all.
(a) $x_{3}=1.4142157,$ within about $0.000002$ of $\sqrt2.$
(b) The iterates run $0,1,0,1,\ldots$ and never approach a root.
Newton's method, step by step
The sketch draws each tangent and fills in a table of the iterates.
- Press step three or four times for $x^{2}-2$ from $x_{0}=2.$ Watch the error column: it does not shrink steadily, it collapses.
- Drag the slider for $x_{0}$ to another starting point and step again. The method still finds $\sqrt2$ in a few steps.
- Press the button with the function until it shows $x^{3}-2x+2,$ and step from $x_{0}=0{:}$ the iterates run $0,1,0,1,\ldots$ for ever.
- On that function, drag $x_{0}$ to the far left of the slider and press step once. The tangent there is nearly flat, and $x_{1}$ lands far outside the picture.
Example 6.6 iterates with numbers. The iteration can also be written once and for all as a formula in $x,$ and for a square root that formula is worth knowing. The next example is the app's own, not one of the notes'.
Newton's iteration for a square root, in general
Let $c>0$ and $f(x)=x^{2}-c.$ Write the Newton iteration $x-\dfrac{f(x)}{f'(x)}$ as simply as possible, and say what it does.
Try it yourself first. Then open one step at a time.
Substitute. $f'(x)=2x,$ so for $x\ne0$ $$ x-\frac{f(x)}{f'(x)}=x-\frac{x^{2}-c}{2x}\,. $$
One fraction. Over the common denominator $2x{:}$ $$ \begin{aligned} x-\frac{x^{2}-c}{2x}&=\frac{2x^{2}-x^{2}+c}{2x}\\ &=\frac{x^{2}+c}{2x}\\ &=\frac12\Bigl(x+\frac{c}{x}\Bigr)\,. \end{aligned} $$ The minus in front of the fraction changes the sign of both terms of $x^{2}-c,$ so $-c$ becomes $+c.$
Read it. If $x$ is too large, then $c/x$ is too small, and the other way round: the new guess is the average of the two. With $c=2$ and $x_{0}=2$ it gives $\tfrac12(2+1)=1.5,$ as in Example 6.6.
Typing it. As an answer to be checked by the app, write
x/2 + c/(2x)with the number in place of $c,$ for examplex/2 + 7/(2x). The brackets matter:7/2xis read as $\tfrac72\,x,$ which is a different function.
$x-\dfrac{x^{2}-c}{2x}=\dfrac12\Bigl(x+\dfrac{c}{x}\Bigr)=\dfrac{x}{2}+\dfrac{c}{2x}{:}$ each step averages the guess $x$ with $c/x.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Which expression is the Newton iteration $x-\dfrac{f(x)}{f'(x)}$ for $f(x)=x^{3}-4,$ simplified?
- Why is that answer tempting? The minus in front of the fraction was applied to $x^{3}$ only. It changes the sign of both terms: $-(x^{3}-4)=-x^{3}+4.$
- Why is that answer tempting? The fraction is upside down: $f'/f$ instead of $f/f'.$ The step is the value divided by the slope — how far to go along the tangent to reach the axis.
- Why is that answer tempting? $\dfrac{x^{3}}{3x^{2}}=\dfrac{x}{3}$ was found and then not subtracted from $x.$ The iteration is $x$ minus the whole fraction, and $x-\dfrac{x}{3}=\dfrac{2x}{3}.$
With $f'(x)=3x^{2},$ $$
\begin{aligned}
x-\frac{x^{3}-4}{3x^{2}}&=\frac{3x^{3}-x^{3}+4}{3x^{2}}\\
&=\frac{2x^{3}+4}{3x^{2}}\\
&=\frac{2x}{3}+\frac{4}{3x^{2}}\,.
\end{aligned}
$$ A check with a number: from $x_{0}=2$ the formula gives $\tfrac43+\tfrac{4}{12}=\tfrac53\approx1.67,$ and directly $2-\tfrac{8-4}{12}=\tfrac53.$ The cube root of $4$ is about $1.59.$ Typed as an answer: 2x/3 + 4/(3x^2).
Existence again: the extreme value theorem
The height of a road is a continuous function of the distance along it. Must a stage of a bicycle race have a highest point?
The same shape, the second time round: a theorem that promises, and then a method that finds. This time what is promised is a largest and a smallest value, and first we need the words for them.
These are what a later lecture will call global extreme points: the comparison is against every point of $S,$ not merely against nearby ones.
Recap Closed, open and bounded intervals
The next Fact speaks of a closed and bounded interval. An interval is closed if it contains every endpoint it has, and bounded if both of its ends are numbers.
| interval | closed | bounded |
|---|---|---|
| $[a,b]$ | yes | yes |
| $(a,b)$ | no | yes |
| $[a,b)$ | no | yes |
| $[a,\infty)$ | yes | no |
A square bracket means that the endpoint belongs to the interval, a round one that it does not; $\infty$ is not a number and is never included. So a closed and bounded interval is one of the form $[a,b],$ with $a$ and $b$ real numbers and both of them included. The open interval $(0,1)$ is not closed: neither $0$ nor $1$ belongs to it.
If $f$ is continuous on a closed and bounded interval $[a,b]$, then $f$ attains both a maximum and a minimum on $[a,b]$.
All three hypotheses are load-bearing. On $[0,1)$ the function $f(x)=x$ climbs towards $1$ and never arrives, so a domain that is not closed can lose the maximum; on $[0,\infty)$ it climbs for ever, so an unbounded domain can too; and a function with a jump can step over its own maximum. A homely version of the extreme value theorem: the height of a road is a continuous function of distance along it, so any stage of a bicycle race has a highest point and a lowest one.
Guess first. The next question is a sentence from an exam script. Decide what is wrong with it before you read the example after it.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
From a student's exam script:
"$f$ is continuous on $(0, 1)$, which is not closed. By the extreme value theorem, $f$ has no maximum."
What is wrong with this?
- Why is that answer tempting? Reads a failed hypothesis as a conclusion. That is the classic misuse of a theorem, and examiners know it well.
- Why is that answer tempting? It is not: neither endpoint belongs to it. A separate confusion about what ‘closed’ means, and worth sorting out (see the box on intervals above).
- Why is that answer tempting? Invents a hypothesis. The extreme value theorem needs continuity only — adding conditions is as wrong as dropping them.
A theorem of the form hypotheses ⟹ conclusion tells you nothing whatever when the hypotheses fail. It does not tell you the conclusion is false.
Both things happen on $(0,1)$:
- $f(x) = x$ has no maximum — it climbs towards $1$ and never arrives
- $f(x) = x(1-x)$ has a maximum, at $x = \tfrac12$
Same interval, same continuity, opposite answers. So "not closed" cannot possibly decide the question.
What the EVT is for is the other direction: on a closed bounded interval you get a guarantee, and can start hunting without first worrying whether there is anything to find.
Not closed is not a verdict
A student writes: “$f$ is continuous on $(0,1),$ which is not closed. By the extreme value theorem, $f$ has no maximum.” What is wrong?
Try it yourself first. Then open one step at a time.
Two continuous functions on that very interval settle it. $f(x)=x$ has no maximum on $(0,1){:}$ it approaches $1$ without reaching it. But $f(x)=x(1-x)$ does have one, at $x=\tfrac12.$ Same interval, same continuity, opposite answers — so “not closed” cannot possibly decide the question, and any argument claiming that it does must be wrong.
What Fact 6.9 is for is the other direction: on a closed bounded interval you have a guarantee, and may start hunting without first worrying whether there is anything to find.
The theorem is read backwards. Its hypotheses failing tells us nothing: on $(0,1)$ the function $x$ has no maximum and $x(1-x)$ has one.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
For which of these does Fact 6.9 guarantee a maximum?
- Why is that answer tempting? The interval is open, so Fact 6.9 says nothing. And here there is in fact no maximum: $x^{2}$ approaches $4$ as $x$ approaches $2,$ but $2$ is not in the interval.
- Why is that answer tempting? This function does have a maximum, $0$ at $x=0.$ But the interval is unbounded, so the guarantee does not come from Fact 6.9. A maximum that happens to exist is not one the theorem promised.
- Why is that answer tempting? The interval is closed and bounded, but $f$ is not continuous at $0.$ And there is no maximum: just to the right of $0$ the values $1/x$ grow without bound.
Only $1/x$ on $[1,4]$ meets all three hypotheses: it is continuous there, and $[1,4]$ is closed and bounded. The maximum is $1,$ at $x=1.$ In each of the other three one hypothesis fails, and then Fact 6.9 is silent — whether or not a maximum exists, as $-x^{2}$ on $[0,\infty)$ shows.
Where to look
A smooth function has its largest value at a point inside an interval. What does the tangent look like there?
Fact 6.9 promises that a maximum exists. Now the method that finds it. Stand at the top of a smooth hill: the ground there is level, and the tangent is horizontal.
The two names mean exactly the same thing. Economists tend to say critical point; both are used in this course.
Let $c$ be an interior point of an interval on which $f$ is defined, let $f$ be differentiable at $c$, and let $c$ be a maximum or minimum point. Then $f'(c)=0$.
The two signs of the quotient. Suppose $c$ is an interior maximum point. For small $h>0$ the quotient $\bigl(f(c+h)-f(c)\bigr)/h$ has a numerator $\le0$ and a positive denominator, so it is $\le0;$ for small $h<0$ the numerator is still $\le0$ but the denominator is negative, so the quotient is $\ge0.$
Both limits are the derivative. Both one-sided limits are $f'(c),$ by Fact 5.8. Hence $f'(c)\le0$ and $f'(c)\ge0$ at once, so $f'(c)=0.$
Both qualifiers in Fact 6.13 do real work. Interior: a maximum at an endpoint need not have a horizontal tangent, and generally does not. Differentiable: a function with a corner can have a minimum exactly at the corner, where there is no derivative to set to zero.
Guess first. The next question asks whether the Fact can be read backwards.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
$f$ is differentiable, $c$ is an interior point of the interval, and $$f'(c) = 0$$
Does $f$ have a maximum or a minimum at $c$?
- Why is that answer tempting? Reads the first-order condition backwards. It is the commonest optimisation error and it survives to the exam unless it is named.
- Why is that answer tempting? The first half is right and the ‘so’ is wrong: f″(c) can be 0, which is exactly what happens for x³. Half a theorem, stated as a whole one.
- Why is that answer tempting? Backwards — the FOC applies at interior points precisely, and endpoints are the case it cannot see.
$f(x) = x^{3}$ has $f'(x) = 3x^{2},$ so $f'(0) = 0$ — and $x^3$ is strictly increasing through $0.$ Not a maximum, not a minimum: it simply pauses.
So the logic runs one way only: $$\text{interior extremum} \;\Longrightarrow\; f'(c) = 0$$ and not back. $f'(c) = 0$ buys you a candidate.
On C. The second-derivative test is real and you will meet it in Lecture 7 — but $f''(0) = 0$ for $x^3,$ and when the test gives $0$ it gives no information at all.
It is worth stopping on that, because this lecture contains three separate traps and they are all the same trap. A theorem says $P\Rightarrow Q.$ There are exactly two ways to misread it: to conclude $Q\Rightarrow P,$ which is the converse, and to conclude $\lnot P\Rightarrow\lnot Q,$ which is the inverse. Reading the first-order condition backwards is the first; concluding from Example 6.11 that a non-closed interval has no maximum is the second. And they are not two errors but one, because $\lnot P\Rightarrow\lnot Q$ is precisely the contrapositive of $Q\Rightarrow P,$ so the two say the same thing. What a theorem's hypotheses buy you is a guarantee when they hold; when they fail, the correct conclusion is silence.
Three functions with a stationary point at $0$ show how little the condition $f'(0)=0$ decides on its own. The example is the app's own.
Three stationary points, three verdicts
Each of $f(x)=x^{2},$ $g(x)=-x^{2}$ and $h(x)=x^{3}$ has a stationary point at $x=0.$ For each, is $0$ a maximum point, a minimum point, or neither, over all of $\mathbb{R}?$
Try it yourself first. Then open one step at a time.
The condition holds for all three. $f'(x)=2x,$ $g'(x)=-2x$ and $h'(x)=3x^{2}$ are all $0$ at $x=0.$
$f(x)=x^{2}.$ For every $x$ we have $f(x)=x^{2}\ge0=f(0),$ so by Definition 6.8 the point $0$ is a minimum point, with minimum value $0.$
$g(x)=-x^{2}.$ For every $x$ we have $g(x)=-x^{2}\le0=g(0){:}$ a maximum point, with maximum value $0.$
$h(x)=x^{3}.$ $h(-1)=-1<0=h(0)$ and $h(1)=1>0=h(0),$ so $0$ is neither: points on one side beat it and points on the other side lose to it (Watch out 6.14).
For $x^{2}$ a minimum point, for $-x^{2}$ a maximum point, and for $x^{3}$ neither. The same condition $f'(0)=0$ holds in all three.
Three kinds of candidate
On the interval $[0,3],$ where does $f(x)=3x^{2}-6x+5$ take its largest value?
So the first-order condition produces a shortlist rather than an answer, and the shortlist has exactly three kinds of entry.
A function shows why. Take $f(x)=x^{3}-3x^{2}+2$ on $[-1.2,\,3.2],$ which has two interior stationary points — and neither of them is the answer. Both global extreme points sit at the ends, where $f'=11.52,$ nowhere near zero. Here $f'(x)=3x^{2}-6x=3x(x-2),$ and the four candidates are:
| $x$ | kind | $f(x)$ |
|---|---|---|
| $-1.2$ | endpoint | $-4.048$ |
| $0$ | $f'(0)=0$ | $2$ |
| $2$ | $f'(2)=0$ | $-2$ |
| $3.2$ | endpoint | $4.048$ |
The minimum is at the left end and the maximum at the right end. The figure below shows the same thing for another function.
Guess first. The next question is the hook of this section. Answer it before the worked example does.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Where does $f$ attain its maximum?
- Why is that answer tempting? Assumes the stationary point is the maximum. Here it is the minimum — f(1) = 2, the smallest of the three candidates.
- Why is that answer tempting? Checks the endpoints but compares them wrongly: f(0) = 5 against f(3) = 14.
- Why is that answer tempting? Believes an extremum must have f′ = 0. Endpoints are candidates precisely because the FOC cannot see them.
$f'(x) = 6x - 6$, zero only at $x = 1$. Candidates: $0$, $1$, $3$.
| $x$ | $f(x)$ | |
|---|---|---|
| $0$ | $5$ | endpoint |
| $1$ | $2$ | minimum |
| $3$ | $14$ | maximum |
The one point where the derivative vanishes is the minimum. The maximum sits at an endpoint, where $f'(3) = 12$ — nowhere near zero.
The parabola opens upward, so of course its vertex is the low point. A thirty-second sketch would have said so before any differentiation.
The only stationary point is the minimum
Find the maximum and minimum of $f(x)=3x^{2}-6x+5$ on $[0,3].$
Try it yourself first. Then open one step at a time.
$f$ is a polynomial, hence continuous, and $[0,3]$ is closed and bounded, so by Fact 6.9 both exist.
It is differentiable everywhere, so there are no candidates of the third kind, and $f'(x)=6x-6=0$ only at $x=1.$ The candidates are therefore $0,$ $1$ and $3{:}$ $$ \begin{aligned} f(0)&=5\,,\\ f(1)&=2\,,\\ f(3)&=14\,. \end{aligned} $$ The maximum value is $14,$ at the endpoint $x=3;$ the minimum value is $2,$ at the stationary point $x=1.$
Notice what happened. The only point where the derivative vanished turned out to be the minimum, and the maximum sat at an endpoint where $f'(3)=12.$ A student who solved $f'=0$ and stopped would have answered the wrong question with confidence. A thirty-second sketch — the parabola opens upwards, so its vertex is the low point — would have said as much before any differentiation.
The maximum value is $14,$ at $x=3,$ and the minimum value is $2,$ at $x=1.$
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
What is the smallest value of $f(x)=x^{2}-6x$ on the interval $[-1,2]?$
- Why is that answer tempting? That is $f(3),$ the value at the stationary point $x=3.$ But $3$ is not in $[-1,2],$ so it is not a candidate.
- Why is that answer tempting? That is $f(-1),$ the largest value on the interval. Both endpoints are candidates, and the comparison picks the smaller one for the minimum.
- Why is that answer tempting? Fact 6.9 guarantees a minimum: $f$ is continuous and $[-1,2]$ is closed and bounded. With no stationary point inside, the minimum is at an endpoint.
$f$ is continuous on the closed and bounded interval $[-1,2],$ so the minimum exists. $f'(x)=2x-6$ is zero only at $x=3,$ outside the interval, so the candidates are the two endpoints: $$ \begin{aligned} f(-1)&=1+6=7\,,\\ f(2)&=4-12=-8\,. \end{aligned} $$ The smallest value is $-8,$ at $x=2.$ (In fact $f'<0$ on the whole interval, so $f$ decreases from $7$ to $-8.$)
When a candidate is the answer
A firm's profit is $12\sqrt{Q}-Q$ for every output $Q>0.$ There are no endpoints to compare. How can it know which output is best?
Method 6.15 decides by comparing numbers, which is reliable and requires a finite list. On an interval that is not closed and bounded there is no such list and Fact 6.9 is silent, so something else must do the work. Two results do, and both look at the sign of $f'$ rather than at values of $f.$
Let $f$ be differentiable on an interval $I$ and let $c\in I$. If $f'(x)\ge0$ for all $x<c$ in $I$ and $f'(x)\le0$ for all $x>c$ in $I$, then $c$ is a maximum point for $f$ over the whole of $I$. If the inequalities are reversed, $c$ is a minimum point.
This needs no endpoints and no comparison: $f$ rises all the way up to $c$ and falls all the way after it, by Fact 2.23, so nothing anywhere in $I$ can beat $f(c).$ What makes it usable is that the sign pattern it asks for is guaranteed by a condition on $f''$ that is often trivial to check.
Before that condition, the first-derivative test on its own, on an interval with no right end. The example is the app's own.
The lowest average cost
Producing $Q$ units costs $C(Q)=Q^{2}+4Q+16,$ so for $Q>0$ the average cost is $$ \begin{aligned} A(Q)&=\frac{C(Q)}{Q}\\ &=Q+4+\frac{16}{Q}\,. \end{aligned} $$ Find the output with the lowest average cost.
Try it yourself first. Then open one step at a time.
No list to compare. The interval $(0,\infty)$ has no endpoints in it, and Fact 6.9 promises nothing there. So we look at the sign of $A'.$
The derivative, factored. $$ \begin{aligned} A'(Q)&=1-\frac{16}{Q^{2}}=\frac{Q^{2}-16}{Q^{2}}\\ &=\frac{(Q-4)(Q+4)}{Q^{2}}\,. \end{aligned} $$ For $Q>0$ the factors $Q+4$ and $Q^{2}$ are positive, so the sign of $A'(Q)$ is the sign of $Q-4.$
The test. $A'(Q)\le0$ for $0<Q<4$ and $A'(Q)\ge0$ for $Q>4.$ By Fact 6.18, with the inequalities reversed and $I=(0,\infty),$ the point $Q=4$ is a minimum point over the whole of $I.$ The lowest average cost is $A(4)=4+4+4=12.$
A check. The marginal cost there is $C'(4)=2\cdot4+4=12,$ equal to the average cost — as Example 3.13 says it must be where the average cost stops falling and starts rising.
$Q=4,$ with average cost $A(4)=12.$
In words: as $\lambda$ runs from $0$ to $1,$ the point $\lambda x+(1-\lambda)y$ runs along the interval from $y$ to $x,$ and $\lambda f(x)+(1-\lambda)f(y)$ runs along the chord from $f(y)$ to $f(x).$ For example $f(x)=-x^{2}$ is concave: with $x=-1,$ $y=1$ and $\lambda=\tfrac12$ the left-hand side is $f(0)=0$ and the right-hand side is $\tfrac12(-1)+\tfrac12(-1)=-1.$ Here the inequality is even strict whenever $x\ne y$ and $0<\lambda<1,$ so $-x^{2}$ is strictly concave, and $x^{2}$ is convex for the same reason.
Let $f$ be twice differentiable on an interval $I$.
- $f$ is concave on $I$ if and only if $f''(x)\le0$ throughout $I$, and convex if and only if $f''(x)\ge0$ throughout. If $f''<0$ throughout then $f$ is strictly concave, though not conversely.
- If $f$ is concave on $I$ and $c$ is an interior point with $f'(c)=0$, then $c$ is a maximum point for $f$ over the whole of $I$. If $f$ is convex, such a $c$ is a minimum point over the whole of $I$.
Part (ii) is the reason concavity matters to economists, and it is a genuinely stronger conclusion than anything in the previous subsection: it delivers a global maximum on an interval of any kind, with no endpoints to check and no list to compare. Where the extreme value theorem is silent, concavity can still speak.
Diminishing returns: one best output
A firm producing $Q>0$ units has profit $\pi(Q) = 12\sqrt{Q} - Q.$ Find the output that maximises profit.
Try it yourself first. Then open one step at a time.
The domain $(0,\infty)$ is neither closed nor bounded, so Fact 6.9 guarantees nothing and Method 6.15 has no finite list to work with. Concavity settles it instead.
Differentiating twice, $$ \begin{aligned} \pi'(Q) &= \frac{6}{\sqrt{Q}} - 1\,,\\ \pi''(Q) &= -3Q^{-3/2} < 0 \end{aligned} $$ for all $Q>0,$ so $\pi$ is concave on the whole domain.
Setting $\pi'(Q)=0$ gives $\sqrt{Q}=6,$ that is $Q=36,$ and by Fact 6.20(ii) this is the global maximum. The maximal profit is $\pi(36) = 72-36 = 36.$
Diminishing returns are what made this work: $\pi''<0$ says each extra unit adds less revenue than the last, and a firm facing diminishing returns has a single best output rather than a shortlist.
The best output is $Q=36,$ with profit $\pi(36)=36.$
Fact 6.20 is the global test. Lecture 7 adds the local one — the second-derivative test at a single point, which asks only for the sign of $f''(c)$ and gives only a local verdict — and Lecture 10 carries both into two variables, where concavity again turns candidates into answers.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
$f$ is twice differentiable and concave on all of $\mathbb{R}.$ Which statement must be true?
- Why is that answer tempting? $f(x)=x$ is concave, since $f''=0\le0,$ and it has no maximum point. Fact 6.20(ii) needs a point with $f'(c)=0,$ and $x$ has none.
- Why is that answer tempting? Concave means $f''\le0$ (Fact 6.20(i)), which allows $f''=0.$ The line $f(x)=x$ is concave with $f''=0$ everywhere.
- Why is that answer tempting? A constant function is concave and has $f'(x)=0$ for every $x,$ and the line $f(x)=x$ has no solution at all. Concavity says what a solution is, not how many there are.
Every point of $\mathbb{R}$ is interior, so if $f'(c)=0$ then Fact 6.20(ii) makes $c$ a maximum point over the whole of $\mathbb{R}.$ That is all concavity promises: whether such a $c$ exists is a separate question, and for $f(x)=x$ the answer is no.
What you should take away
- Intermediate value theorem: $f$ continuous on $[a,b]$ with a sign change has a root in $(a,b).$ Continuity is the whole hypothesis ($1/x$ on $[-1,1]$), and the conclusion is existence only. Adding strict monotonicity turns “at least one” into “exactly one” — two separate arguments for two separate jobs.
- Newton's method: $x_{n} = x_{n-1}-f(x_{n-1})/f'(x_{n-1}),$ which is the tangent line solved for its own root, iterated. Usually very fast — the correct digits roughly double each step — and it can also cycle for ever.
- Extreme value theorem: $f$ continuous on a closed bounded interval attains a maximum and a minimum. All three hypotheses are load-bearing, and they are sufficient, not necessary.
- First-order condition: an interior extreme point at which $f$ is differentiable has $f'(c)=0.$ Necessary, not sufficient — recall $x^{3}$ at $0.$
- Reading the first-order condition backwards and reading the extreme value theorem backwards are the same error, since $\lnot P\Rightarrow\lnot Q$ is the contrapositive of $Q\Rightarrow P.$ When hypotheses fail, a theorem says nothing.
- To find extrema on $[a,b]{:}$ list the candidates — interior stationary points, endpoints, and interior points where $f'$ fails to exist — evaluate, and compare. The comparison decides; the derivative only shortlists.
- Where there is no finite list, sign conditions do the work. If $f'$ is $\ge0$ before $c$ and $\le0$ after it, $c$ is a maximum over the whole interval; and $f''\le0$ throughout means $f$ is concave, which makes every interior stationary point a global maximum.
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — what a sign change needs
$f$ satisfies $f(-1)=-1$ and $f(1)=1.$ Must there be a $c$ in $(-1,1)$ with $f(c)=0?$
(A) yes, always;
(B) yes, provided $f$ is continuous on $[-1,1];$
(C) yes, provided $f$ is differentiable on $(-1,1);$
(D) no — a sign change never guarantees a root.
One way to think about it
(B), by Fact 6.1. (A) fails for $1/x,$ which is not continuous at $0$ (Example 6.3). (D) is too pessimistic: with continuity the root is guaranteed.
(C) fails too, and the counterexample is worth seeing. Let $f(-1)=-1$ and $f(x)=1$ for $-1<x\le1.$ On the open interval $(-1,1)$ this $f$ is constant, so it is differentiable there, and it is never zero in $(-1,1).$ What goes wrong is at the endpoint $x=-1,$ where $f$ jumps: differentiability on the open interval says nothing about the ends.
Discussion question 2 — one step of Newton
Apply one step of Newton's method to $f(x)=x^{2}-2$ from $x_{0}=2.$
(A) $1.5;$
(B) $2.5;$
(C) $1.0;$
(D) $1.4142.$
Which wrong answer is a sign of having understood where the method is going?
One way to think about it
(A): $x_{1}=2-\dfrac{f(2)}{f'(2)}=2-\dfrac{2}{4}=1.5.$ (B) adds the step instead of subtracting it, and (C) divides by $x_{0}=2$ instead of by the slope $f'(2)=4.$
(D) is the wrong answer that shows understanding: $1.4142$ is $\sqrt2,$ the root the iteration is heading for. Only the number of steps is wrong. One step lands where the tangent meets the axis, and that is $1.5;$ it takes three steps to reach $1.4142157.$
Discussion question 3 — a horizontal tangent that is not an extremum
$f$ is differentiable, $c$ is interior, and $f'(c)=0.$ Does $f$ have an extreme point at $c?$ Give the standard counterexample, and say which lecture you first met it in.
One way to think about it
Not necessarily. The standard counterexample is $f(x)=x^{3}$ at $c=0{:}$ $f'(0)=0,$ but $f(-1)=-1<0<1=f(1),$ so $0$ is neither a maximum nor a minimum point. It was first met in Lecture 2, in Watch out 2.24, where it showed that a strictly increasing function need not have $f'>0$ everywhere. Here it shows that Fact 6.13 cannot be read backwards.
Discussion question 4 — the same function on a shorter interval
In Example 6.17, suppose the interval were $[0,1]$ instead. Where are the maximum and minimum then, and which step of Method 6.15 changes?
One way to think about it
Step 2, the list of candidates, changes. The point $3$ is no longer in the interval, and the stationary point $x=1$ is now an endpoint as well. The candidates are $0$ and $1,$ with $f(0)=5$ and $f(1)=2.$ The maximum is $5,$ at $x=0,$ and the minimum is $2,$ at $x=1.$
Step 1 is unchanged ($[0,1]$ is closed and bounded), and step 3 still decides. In fact $f'(x)=6x-6\le0$ on $[0,1],$ so $f$ decreases from $5$ to $2$ all the way.
Discussion question 5 — concave, and no maximum
Give a function that is concave on ${\mathbb R}$ with no maximum point at all. Why does this not contradict Fact 6.20(ii)?
One way to think about it
$f(x)=x$ will do: $f''(x)=0\le0,$ so it is concave by Fact 6.20(i), and it grows without bound. A strictly concave one is $f(x)=-e^{x}{:}$ $f''(x)=-e^{x}<0,$ and its values approach $0$ from below without ever reaching it.
No contradiction, because part (ii) has a hypothesis: an interior point $c$ with $f'(c)=0.$ Neither function has one ($f'(x)=1$ and $f'(x)=-e^{x}$ are never zero), so the Fact is silent — the same lesson as Watch out 6.10.
Discussion question 6 — strictly convex with f″(0) = 0
Is $f(x)=x^{4}$ strictly convex? What is $f''(0)?$ Which half of Fact 6.20(i) does this illustrate?
One way to think about it
Yes, $x^{4}$ is strictly convex: every chord lies strictly above the graph between its two ends. For instance the chord from $(-1,1)$ to $(1,1)$ is the line $y=1,$ and $x^{4}<1$ for $-1<x<1.$ (In general, the slope $f'(x)=4x^{3}$ is strictly increasing, so the graph bends upwards everywhere and never runs straight.)
But $f''(x)=12x^{2},$ so $f''(0)=0.$ This is the “though not conversely” of part (i): $f''>0$ throughout makes a function strictly convex, but a strictly convex function need not have $f''>0$ everywhere. Compare $x^{3},$ which is strictly increasing with $f'(0)=0.$
Discussion question 7 — when the hypothesis is false
A theorem says $P\Rightarrow Q$ and you observe that $P$ is false. What may you conclude about $Q?$
One way to think about it
Nothing. The theorem promises $Q$ only when $P$ holds; when $P$ fails, $Q$ may be true or false. Concluding “not $Q$” is the inverse, $\lnot P\Rightarrow\lnot Q,$ which is the contrapositive of the converse $Q\Rightarrow P$ — the same error as reading the theorem backwards.
Example 6.11 shows both outcomes. On $(0,1),$ where the hypothesis of the extreme value theorem fails, $x(1-x)$ has a maximum and $x$ has none.