Question

The graph of a function \(f\) is shown.

(a) Find the average rate of change of \(f\)on the interval \(\left( {20,60} \right)\).

(b) Identify an interval on which the average rate of change of \(f\)is \(0\).

(c) Compute

\(\frac{{f\left( {40} \right) - f\left( {10} \right)}}{{40 - 10}}\)

What does this value represent geometrically?

(d) Estimate the value of \(f'\left( {50} \right)\).

(e) Is \(f'\left( {10} \right) > f'\left( {30} \right)\)?

(f ) Is \(f'\left( {60} \right) > \frac{{f\left( {80} \right) - f\left( {40} \right)}}{{80 - 40}}\)? Explain

Short answer

(a) \(\frac{{f\left( {60} \right) - f\left( {20} \right)}}{{60 - 20}} = 10\)on \(\left( {20,60} \right)\).

(b) The average rate of change of \(f\) is zero on \(\left( {10,50} \right)\).

(c) \(\frac{{f\left( {40} \right) - f\left( {10} \right)}}{{40 - 10}} = - \frac{{20}}{3}\).

(d) \(f'\left( {50} \right) = 25\).

(e) No

(f) Yes

Step-by-step solution

(a) Step 1: Finding average rate of change of a function on an interval 

The average rate of changeof \(f\) on \(\left( {20,60} \right)\) is:

\(\begin{aligned}\frac{{f\left( {60} \right) - f\left( {20} \right)}}{{60 - 20}} &= \frac{{700 - 300}}{{40}}\\ &= \frac{{400}}{{40}}\\ &= 10\end{aligned}\)

Hence average rate of change of the function is \(10\).

(b) Step 2: Interval on which average rate of change is zero

The average rate of change will zero when at two points they have same functional value. Considering the interval \(\left( {10,50} \right)\). The average rate of change on this interval will be \(\frac{{f\left( {50} \right) - f\left( {10} \right)}}{{50 - 10}} = \frac{{400 - 400}}{{40}} = 0\). So this a required interval on which average rate of change is zero.

(c)  Step 3: Computing the given expression

Computing the given expression we get

\(\begin{aligned}\frac{{f\left( {40} \right) - f\left( {10} \right)}}{{40 - 10}} &= \frac{{200 - 400}}{{30}}\\ &= \frac{{ - 200}}{{30}}\\ &= - \frac{{20}}{3}\end{aligned}\)

This is the slope at the point of the function which lies between the points \(\left( {10,f\left( {10} \right)} \right)\) and \(\left( {40,f\left( {40} \right)} \right)\).

(d)  Step 4: Estimate the slope at a given point

From the graph it seems the slope at \(\left( {50,f\left( {50} \right)} \right)\) is almost the slope of the line segment between the points \(\left( {40,f\left( {40} \right)} \right)\)and \(\left( {60,f\left( {60} \right)} \right)\). Hence the slope will be:

\(\begin{aligned}f'\left( {50} \right) \approx \frac{{f\left( {60} \right) - f\left( {40} \right)}}{{60 - 40}}\\ \approx \frac{{700 - 200}}{{20}}\\ \approx 25\end{aligned}\)

Hence \(f'\left( {50} \right) \approx 25\)

(e) Step 5: Comparing derivative of the function at two points

The slope at the point \(10\) is more steeper than the slope at the point \(30\). But since at both of the point the slope is negative and hence \(f'\left( {10} \right) < f'\left( {30} \right)\).

Comparing the slope at a point and the slope of a line segment

From the graph we can observe that slope at the point \(60\) is greater than the slope of the line segment between the points \(\left( {40,f\left( {40} \right)} \right)\) and \(\left( {80,f\left( {80} \right)} \right)\).

Hence the statement is true, so \(f'\left( {60} \right) > \frac{{f\left( {80} \right) - f\left( {40} \right)}}{{80 - 40}}\).