MAT147 — Week 3

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Week 3 — Review Questions

Units, rates of change, reading graphs of derivatives, tangent lines, and differentiability

Question 1 — Interpreting an Inverse

Let $V = f(t)$ be the number of liters of water in a tank $t$ minutes after it starts draining.

Explain what $f^{-1}(50) = 8$ means in context.

Solution 1

The inverse function $f^{-1}$ takes an amount of water (in liters) as input and gives a time (in minutes) as output.

So $f^{-1}(50) = 8$ means that 8 minutes after the tank starts draining, it contains 50 liters of water. Equivalently, $f(8) = 50\text{.}$

Question 2 — Average Velocity

A particle’s position is given by

\begin{equation*} s(t) = t^2 + 4t \end{equation*}

where $s$ is in meters and $t$ is in seconds. Find the average velocity on the interval $[2, 3]\text{.}$

Solution 2

Average velocity is the change in position divided by the change in time:

\begin{equation*} s(3) = 9 + 12 = 21, \qquad s(2) = 4 + 8 = 12 \end{equation*}

\begin{equation*} \dfrac{s(3) - s(2)}{3 - 2} = \dfrac{21 - 12}{1} = 9 \text{ m/s} \end{equation*}

Question 3 — Signs of $P'$ and $P''$

Let $P(t)$ be the population of a town at time $t\text{.}$ For each statement, give the signs of $P'(t)$ and $P''(t)\text{.}$

The population is growing, but more and more slowly.
The population is decreasing faster and faster.

Solution 3

$P'$ tells whether the population is growing or shrinking. $P''$ tells whether that rate is increasing or decreasing.

(a) Growing means $P'(t) \gt 0\text{.}$ Growing more slowly means the rate is decreasing, so $P''(t) \lt 0\text{.}$

(b) Decreasing means $P'(t) \lt 0\text{.}$ Decreasing faster and faster means the rate is becoming more negative, so $P''(t) \lt 0\text{.}$

Question 4 — Reading a Graph of $f'(x)$

The graph of $f'(x)$ (not $f(x)$) is shown.

Where does $f$ have a local maximum?
Where is $f''(x) = 0\text{?}$

Graph of f prime on the interval from x=-3 to x=7. The curve is negative for x less than -2, crosses zero at x=-2, rises to a local maximum of about 2.35 near x=-0.57, passes through (0,2), crosses zero at x=1, falls to a local minimum of about -3.9 at x=3, rises to touch zero at x=5 without crossing, then falls again.

Solution 4

The graph shown is $f'(x)\text{,}$ so its sign tells where $f$ increases or decreases, and its slope is $f''(x)\text{.}$

(a) $f$ has a local maximum where $f'$ changes from positive to negative. That happens only at $x = 1$. At $x = 5\text{,}$ $f'$ touches zero but stays negative on both sides, so $f$ has no local extremum there.

(b) $f''(x) = 0$ where the graph of $f'$ has a horizontal tangent line, at its peaks and valleys: $x \approx -0.57\text{,}$ $x = 3\text{,}$ and $x = 5$.

Question 5 — Matching $f\text{,}$ $f'\text{,}$ and $f''$

Graphs A, B, and C show $f(x)\text{,}$ $f'(x)\text{,}$ and $f''(x)\text{.}$ Identify which graph is which, and give a reason.

$f(x)$ is graph ____
$f'(x)$ is graph ____
$f''(x)$ is graph ____

Three curves on the interval from x=-5 to x=5, with y from -2 to 2. Curve A (blue) is a wave through the origin with a high near x=-1.4 and a low near x=1.4. Curve B (red) is a wavy S-shaped curve through the origin with a low near x=-1.8 and a high near x=1.8. Curve C (green) is a W-like curve equal to 1 at x=0, dipping below the axis near x=-2.7 and x=2.7.

Solution 5

A derivative is zero wherever the original function has a peak or valley.

Graph B has peaks and valleys at $x \approx \pm 1.8$ and $x \approx \pm 3.5\text{,}$ and graph C crosses zero at exactly those points. So C is the derivative of B.

Graph C has its lowest points at $x \approx \pm 2.7\text{,}$ and graph A crosses zero at exactly those points. So A is the derivative of C.

(a) $f(x)$ is graph B. (b) $f'(x)$ is graph C. (c) $f''(x)$ is graph A.

Question 6 — Average Acceleration from a Table

A car’s velocity is given in the table below.

$t$ (s)	0	1	2	3
$v(t)$ (ft/s)	0	22	38	50

Find the average acceleration over the first second and over the second second. Include units.

Solution 6

Average acceleration is the change in velocity divided by the change in time.

First second ($t = 0$ to $t = 1$):

\begin{equation*} \dfrac{v(1) - v(0)}{1 - 0} = \dfrac{22 - 0}{1} = 22 \text{ ft/s}^2 \end{equation*}

Second second ($t = 1$ to $t = 2$):

\begin{equation*} \dfrac{v(2) - v(1)}{2 - 1} = \dfrac{38 - 22}{1} = 16 \text{ ft/s}^2 \end{equation*}

Question 7 — Slope of a Tangent Line

The tangent line to $y = f(x)$ at the point $(2, 5)$ passes through the point $(6, 17)\text{.}$

Find $f'(2)\text{.}$

Solution 7

$f'(2)$ is the slope of the tangent line at $x = 2\text{.}$ The line passes through $(2, 5)$ and $(6, 17)\text{,}$ so

\begin{equation*} f'(2) = \dfrac{17 - 5}{6 - 2} = \dfrac{12}{4} = 3 \end{equation*}

Question 8 — Differentiability

Let

\begin{equation*} f(x) = |4x - 12| \end{equation*}

Find all $x$ where $f(x)$ is not differentiable, and explain why.

Solution 8

The expression inside the absolute value is zero when $4x - 12 = 0\text{,}$ so at $x = 3\text{.}$

For $x \lt 3\text{,}$ $f(x) = -(4x - 12)\text{,}$ which has slope $-4\text{.}$ For $x \gt 3\text{,}$ $f(x) = 4x - 12\text{,}$ which has slope $4\text{.}$

The left- and right-hand slopes disagree, so the graph has a sharp corner at $x = 3\text{.}$ Therefore $f$ is not differentiable at $x = 3$.