The Derivative and Its Definition: practice solutions, College

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Practice solutions College : The Derivative and Its Definition — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

1 Average rate of change ★★★

$f(1)=2$ and $f(4)=17$. The average rate is $\dfrac{f(4)-f(1)}{4-1}=\dfrac{15}{3}=5$.

Answer: 5.

2 A linear difference quotient ★★★

$f(x+h)-f(x)=5(x+h)-2-(5x-2)=5h$. So the quotient is $\dfrac{5h}{h}=5$ for every $h\neq 0$, and $f'(x)=5$. A line has the same slope everywhere.

3 Power rule practice ★★★

(a) $7x^6$. (b) $10x^9$. (c) $6\cdot 4x^3=24x^3$, using the constant multiple rule.

4 A quadratic polynomial ★★★

Term by term: $f'(x)=3\cdot 2x-8\cdot 1+0=6x-8$.

5 Skateboard distance ★★★

$s(1)=4$ and $s(3)=36$. Average velocity $=\dfrac{36-4}{3-1}=16$ ft/s.

In meters: $16\times 0.3048\approx 4.88$ m/s.

Answer: 16 ft/s (about 4.88 m/s).

6 True or false? ★★★

(a) False: the derivative of a constant is $0$. (b) True: constant multiple rule, $5\cdot 1=5$. (c) False: the exponent drops by one, $4x^3$. (d) False: $|x|$ is continuous at $0$ but not differentiable there.

7 Slope of a tangent ★★★

$f'(x)=2x$, so the slope is $f'(3)=6$. The point is $(3,9)$. The tangent is $y=9+6(x-3)=6x-9$.

8 Derivative at a point from the definition ★★★

$f(2)=-4$. $f(2+h)=4+4h+h^2-8-4h=h^2-4$. So $f(2+h)-f(2)=h^2$ and the quotient is $h$. Its limit as $h\to 0$ is $0$.

Answer: $f'(2)=0$ (horizontal tangent at the vertex).

9 Tangent line to a cubic ★★★

$f(1)=1$. $f'(x)=6x^2-1$, so $f'(1)=5$. Tangent: $y=1+5(x-1)=5x-4$.

10 Temperature during the day ★★★

(a) $T'(t)=-t+6$.

(b) $T'(2)=4$: at $t=2$ the temperature is rising at 4 °F per hour. $T'(8)=-2$: at $t=8$ it is falling at 2 °F per hour.

(c) $T'(t)=0$ at $t=6$, where the rate changes from positive to negative. $T(6)=-18+36+50=68$ °F, and $\dfrac{5}{9}(68-32)=20$ °C.

11 Horizontal tangents ★★★

A horizontal tangent means $f'(x)=0$. $f'(x)=3x^2-12=3(x-2)(x+2)$, so $x=2$ or $x=-2$.

$f(2)=8-24=-16$ and $f(-2)=-8+24=16$.

Answer: the points $(2,-16)$ and $(-2,16)$, as in the figure.

12 Powers and roots ★★★

Rewrite $f(x)=x^{-2}+3x^{1/2}$. Then $f'(x)=-2x^{-3}+\dfrac32x^{-1/2}=-\dfrac{2}{x^3}+\dfrac{3}{2\sqrt{x}}$.

At $x=4$: $-\dfrac{2}{64}+\dfrac{3}{4}=-\dfrac{1}{32}+\dfrac{24}{32}=\dfrac{23}{32}$.

13 A corner ★★★

$f(2)=0$, so the quotient is $\dfrac{|h|}{h}$. For $h>0$ it equals $1$; for $h<0$ it equals $-1$. The one-sided limits are $1$ and $-1$; they differ, so the limit does not exist.

Conclusion: $f'(2)$ does not exist (the graph has a corner at $(2,0)$).

14 Where is the slope 4? ★★★

$f'(x)=2x-6$. Solve $2x-6=4$: $x=5$. Then $f(5)=25-30+1=-4$.

Answer: the point $(5,-4)$.

15 Definition applied to a reciprocal ★★★

$\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=\dfrac{x-(x+h)}{h\,x(x+h)}=\dfrac{-h}{h\,x(x+h)}=\dfrac{-1}{x(x+h)}$. As $h\to 0$ this tends to $-\dfrac{1}{x^2}$.

So $f'(x)=-\dfrac{1}{x^2}$, consistent with the power rule for $x^{-1}$. $f'(2)=-\dfrac14$, $f(2)=\dfrac12$.

Tangent: $y=\dfrac12-\dfrac14(x-2)=-\dfrac14x+1$.

16 Making a piecewise function differentiable ★★★

Continuity is needed: $f(1)=1$, so $a+b=1$.

Equal slopes: the left derivative is $(x^2)'=2x=2$ at $1$; the right derivative is $a$. So $a=2$.

Then $b=1-2=-1$.

Answer: $a=2$, $b=-1$, i.e. $f(x)=2x-1$ for $x>1$.

17 Marginal cost ★★★

(a) $C'(q)=0.04q+3$.

(b) $C'(100)=4+3=7$ dollars per unit.

$C(100)=200+300+500=1000$ and $C(101)=0.02\cdot 10201+303+500=204.02+303+500=1007.02$. The extra cost of the 101st unit is $\$7.02$, very close to $C'(100)=\$7$.

18 Tangent lines through an outside point ★★★

The tangent at $x=a$ is $y=a^2+2a(x-a)=2ax-a^2$. It passes through $(0,-4)$ when $-a^2=-4$, so $a=2$ or $a=-2$.

Answer: $y=4x-4$ (tangent at $(2,4)$) and $y=-4x-4$ (tangent at $(-2,4)$).

19 Parallel tangents ★★★

Parallel means slope $12$. $3x^2=12$ gives $x=\pm 2$.

At $x=2$: point $(2,8)$, tangent $y=8+12(x-2)=12x-16$. At $x=-2$: point $(-2,-8)$, tangent $y=-8+12(x+2)=12x+16$.

20 A ball thrown upward ★★★

(a) $v(t)=-32t+96$.

(b) $v(1)=64$ ft/s $\approx 19.5$ m/s.

(d) $v(5)=-160+96=-64$ ft/s. Negative: the ball is on its way down at 64 ft/s.

21 A vertical tangent ★★★

$f$ is continuous at $0$ because cube roots are continuous. The quotient is $\dfrac{h^{1/3}-0}{h}=h^{-2/3}=\dfrac{1}{h^{2/3}}$, which grows without bound as $h\to 0$ (for $h=0.001$ it equals $100$; for $h=0.000001$ it equals $10{,}000$). The limit is infinite, so $f'(0)$ does not exist.

Conclusion: the graph has a vertical tangent (the $y$-axis) at the origin.