Powers, exponentials, logarithms, and the derivative

(a) Read the exponent. $Y$ is a power function with exponent $r = \tfrac13,$ and $0 < \tfrac13 < 1.$ By Fact 2.2 the graph increases and bends downwards. The factor $12$ stretches it vertically and does not change the shape.

(b) Compare, and let $K$ cancel. $$ \begin{aligned} \frac{12\,(8K)^{1/3}}{12\,K^{1/3}} &= 8^{1/3} = 2\,,\\ \frac{12\,(27K)^{1/3}}{12\,K^{1/3}} &= 27^{1/3} = 3\,. \end{aligned} $$ Eight times the capital doubles output, and $27$ times the capital triples it, whatever $K$ was to begin with.

(c) The numbers. $Y(8) = 12\cdot 2 = 24,$ $Y(27) = 12\cdot 3 = 36$ and $Y(64) = 12\cdot 4 = 48.$ The first extra $12$ units needed $27 - 8 = 19$ more units of capital, the next $12$ units needed $64 - 27 = 37$ more.

In words. Each further unit of capital adds less output than the one before: diminishing returns. It follows from the single fact that the exponent is below $1.$

(a) Substitute. Put $u = e^{x},$ so that $e^{-x} = 1/u$ and $u > 0.$ The equation becomes $u - 10/u = 3.$ Multiply by $u{:}$ $$ \begin{aligned} &u^{2} - 3u - 10 = 0\\ &\Longleftrightarrow\quad (u - 5)(u + 2) = 0\,. \end{aligned} $$

Reject what cannot be. $u = -2$ is impossible, since $u = e^{x}$ is always positive. So $u = 5,$ that is $e^{x} = 5$ and $x = \ln 5.$

Check. $e^{\ln 5} - 10e^{-\ln 5} = 5 - \tfrac{10}{5} = 3.$

(b) Take $\ln$ of both sides. Both sides are positive, and a product becomes a sum: $$ \ln 3 + x\ln 2 = \ln 7 + x\ln 5\,. $$

Solve the linear equation. Collect the terms with $x{:}$ $x(\ln 2 - \ln 5) = \ln 7 - \ln 3,$ so $$ \begin{aligned} x &= \frac{\ln 7 - \ln 3}{\ln 2 - \ln 5}\\ &= \frac{\ln(7/3)}{\ln(2/5)} \approx -0.92\,. \end{aligned} $$ Check with a calculator: $3\cdot 2^{-0.9247}\approx 1.58$ and $7\cdot 5^{-0.9247}\approx 1.58.$

(a) The equation. Growing by the factor $k$ from the moment $t$ to the moment $t + T$ means $f(t + T) = k\,f(t),$ that is $$ A\,a^{t}a^{T} = k\,A\,a^{t}\,. $$

Cancel. $A\,a^{t}$ is never zero, so we may divide by it: $a^{T} = k.$ Neither $A$ nor $t$ is left in the equation, so $T$ depends on neither.

Take ${\ln}.$ $T\ln a = \ln k,$ and $\ln a \ne 0$ because $a > 1,$ so $T = \ln k/\ln a.$ (Since $\ln a > 0$ and $\ln k > 0,$ the time is positive, as it should be.)

(b) Divide. With $k = 3$ and $k = 2,$ $$ \frac{T_{3}}{T_{2}} = \frac{\ln 3/\ln a}{\ln 2/\ln a} = \frac{\ln 3}{\ln 2}\,. $$ The base has cancelled: the ratio is the same for every exponential growth, about $1.585.$

(c) $T_{3} = 9\cdot\dfrac{\ln 3}{\ln 2} \approx 9\cdot 1.585 \approx 14.3$ years.

(a) Step 1: expand. $f(a+h) = 2(a+h)^{2} - 5(a+h) + 1 = 2a^{2} + 4ah + 2h^{2} - 5a - 5h + 1.$

Step 2: the difference. Subtract $f(a) = 2a^{2} - 5a + 1{:}$ $$ \begin{aligned} &f(a+h) - f(a)\\ &\quad= 4ah + 2h^{2} - 5h\,. \end{aligned} $$ Everything without $h$ has cancelled, as it must.

Steps 3 and 4: divide by $h$ and cancel. $$ \begin{aligned} &\frac{f(a+h)-f(a)}{h}\\ &\quad= \frac{h(4a + 2h - 5)}{h}\\ &\quad= 4a - 5 + 2h\,. \end{aligned} $$

Step 5: let $h \to 0.$ The quotient tends to $4a - 5,$ so $f'(a) = 4a - 5.$ This agrees with Fact 2.20: $2\cdot 2\cdot a + (-5).$

(b) In step 4, when the factor $h$ was cancelled: that is division by $h,$ which is allowed only for $h \ne 0.$ At $h = 0$ the Newton quotient is $0/0,$ which is not a number: there is no secant through a point and itself. So the order matters: cancel first, then let $h \to 0.$

(c) The tangent is horizontal where its slope is $0{:}$ $4a - 5 = 0,$ so $a = \tfrac54.$ (That is the vertex of the parabola, $-b/2a$ with the coefficients $2$ and $-5.)$

(a) $f(0) = 0,$ so the Newton quotient at $0$ is $$ \frac{|0 + h| - |0|}{h} = \frac{|h|}{h}\,. $$ For $h > 0$ it is $h/h = 1.$ For $h < 0$ it is $-h/h = -1.$

(b) The derivative is the one value the quotient settles on as $h \to 0$ (Definition 2.17). Here it is $1$ for every positive $h$ and $-1$ for every negative $h,$ however small. There is no single value that it approaches, so the limit does not exist and $f$ is not differentiable at $0.$

In the picture. Secants to the right of the corner all have slope $1,$ secants to the left all have slope $-1.$ They do not settle on one line: this is the sharp corner of Watch out 2.16.

(c) For $|h| < 2$ the number $2 + h$ is positive, so $|2 + h| = 2 + h,$ and $$ \frac{|2+h| - |2|}{h} = \frac{2 + h - 2}{h} = 1\,. $$ The quotient is $1$ for every small $h \ne 0,$ of either sign, so it tends to $1{:}$ $f'(2) = 1.$ Near $x = 2$ the graph is the line $y = x.$

(a) By Fact 2.20 with $a = 0.02,$ $b = 8,$ $c = 500{:}$ $C'(Q) = 0.04\,Q + 8,$ so $C'(200) = 16.$ One more unit costs about $16.$

(b) $C(201) - C(200) = 8 + 0.02\,(201^{2} - 200^{2}) = 8 + 0.02\cdot 401 = 16.02.$ By Fact 2.21, $C(a+h) - C(a) \approx C'(a)\cdot h$ for small $h;$ with $h = 1$ this is $16.$ The difference $0.02$ is the term $0.02\,h^{2}$ that the approximation leaves out.

(c) The average rate of change is the Newton quotient with $h = 50{:}$ $$ \begin{aligned} &\frac{C(250) - C(200)}{50}\\ &= \frac{50\cdot 8 + 0.02\,(250^{2} - 200^{2})}{50}\\ &= \frac{400 + 450}{50} = 17\,. \end{aligned} $$ It is larger than $C'(200) = 16$ because the marginal cost rises with $Q{:}$ over the $50$ units it runs from $16$ up to $C'(250) = 18.$ The average over a long stretch is not the rate at its starting point.

(d) Nothing. The constant cancels in every difference and has derivative $0{:}$ a fixed cost cannot affect the cost of one more unit.

(a) The sign diagram. The zeros of $f'$ are $x = 0$ and $x = 2.$

Report. $f' \le 0$ on $(-\infty, 0]$ and $f' \ge 0$ on $[0, \infty).$ By Fact 2.23, $f$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty).$

(b) False. $f'$ is zero at $2$ but does not change sign there: it is positive on both sides, because the factor $(x-2)^{2}$ is a square. The graph has a horizontal tangent at $x = 2$ and goes on rising, as $x^{3}$ does at $0.$ The graph turns round at $x = 0,$ where $f'$ changes from negative to positive.

(c) Suppose not. By (a), $f$ is increasing on $[0, \infty).$ If it were not strictly increasing, there would be two points $x_1 < x_2$ in $[0, \infty)$ with $f(x_1) = f(x_2).$

Then $f$ is constant between them. For every $x$ between $x_1$ and $x_2$ we have $f(x_1) \le f(x) \le f(x_2),$ because $f$ is increasing, and the two ends are equal. So $f$ is constant on $[x_1, x_2].$

Fact 2.23 gives a contradiction. A function that is constant on an interval has $f'(x) = 0$ at every point of it. But $f'$ is zero only at $x = 0$ and $x = 2,$ not on a whole interval. So there are no such points, and $f$ is strictly increasing on $[0, \infty).$