Question

For the graph of f shown next at the left, list the following quantities in order from least to greatest:

(a) the average rate of change of f on$[-1,1]$

(b) the instantaneous rate of change of f at$x=1$

(c) $f\text{'}(-1)$

(d) $\frac{f\left(2\right)-f(-1)}{2-(-1)}$

Short answer

(a) The average rate of change of the function on $[-1,1]$is $1$

(b) The instantaneous rate of change of the function of the function at $x=1$is $1$

(c) The value of $f(-1)$is $-1$

(d) $\frac{f\left(2\right)-f(-1)}{2-(-1)}=0$

From the values, the quantities in order from the least to the greatest is $\left(c\right),\left(d\right),\left(a\right),\left(b\right)$

Step-by-step solution

Part (a) Step 1. Given Information.

The graph of the function.

Part (a) Step 2. Draw a secant line.

Draw a secant line between the points $x=-1tox=1$

Part (a) Step 3. Calculate average rate of change.

Average rate of change$=\frac{f\left(1\right)-f(-1)}{1-(-1)}$

$$\begin{gathered} =\frac{1-(-1)}{1+1} \\ =\frac{2}{2} \\ =1 \end{gathered}$$

Part (b) Step 1. Draw a tangent line.

Draw a tangent line at the point $x=-1$.

Part (b) Step 2. Find average rate of change.

Consider the interval $[1,1.5]$

The average rate of change in this interval is:

$$\begin{gathered} \frac{f(1.5)-f\left(1\right)}{1.5-1}=\frac{0.9-1}{1.5-1} \\ =\frac{-0.1}{0.5} \\ =-0.2 \end{gathered}$$

Consider the interval $[1,1.1]$

$$\begin{gathered} \frac{f(1.1)-f\left(1\right)}{1.1-1}=\frac{1.1-1}{1.1-1} \\ =\frac{0.1}{0.1} \\ =1 \end{gathered}$$

The instantaneous rate of change of function at $x=1$ is $1$.

Part (c) Step 1. Find f'(-1)

When $x=-1$, the derivative is,

Consider the interval $[-1,-0.5]$,

$$\begin{gathered} f\text{'}(-1)=\frac{f(-0.5)-f(-1)}{-0.5-(-1)} \\ =\frac{-1.5-(-1)}{0.5} \\ =-1 \end{gathered}$$

Part (d) Step 1. Find the given derivative.

Substitute the function values from the graph,

$$\begin{gathered} \frac{f\left(2\right)-f(-1)}{2-(-1)}=\frac{-1-(-1)}{2+1} \\ =\frac{0}{3} \\ =0 \end{gathered}$$