Question

drag the purple x on the graph below to vary the length of the interval, h.

a. use the applet above to determine the value of the difference quotient (the average rate of change) near the point (2, f(2)) when:
i. the length of h is 5.
ii. the length of h is 2.
iii. the length of h is 1.
iv. the length of h is 0.4.
v. the length of h is 0.1.
b. determine if the following statement is true or false.
as the value of h gets smaller and smaller, the value of the difference quotient becomes a better and better approximation of the average rate of change of a function f near the point (2, f(2)).

Explanation:

Step1: Recall the difference - quotient formula

The difference - quotient formula for the average rate of change of a function $y = f(x)$ over the interval $[a,a + h]$ is $\frac{f(a + h)-f(a)}{h}$. Here $a = 2$, so we need to find $f(2 + h)$ and $f(2)$ from the graph. Given $f(2)=9.60$.

Step2: When $h = 5$

The point $(2 + h,f(2 + h))=(7,f(7))$. From the graph, if we assume a linear - like behavior (since we are using the average rate of change), we find the value of $y$ at $x = 7$. Let's assume we can estimate the value of $f(7)$. The average rate of change $\frac{f(2 + 5)-f(2)}{5}=\frac{f(7)-9.60}{5}$. If we estimate $f(7)$ from the graph (say $f(7)=48$), then $\frac{48 - 9.60}{5}=\frac{38.4}{5}=7.68$.

Step3: When $h = 2$

The point $(2 + h,f(2 + h))=(4,f(4))$. Estimating from the graph, if $f(4)=24$, then the average rate of change $\frac{f(4)-f(2)}{2}=\frac{24 - 9.60}{2}=\frac{14.4}{2}=7.2$.

Step4: When $h = 1$

The point $(2 + h,f(2 + h))=(3,f(3))$. Estimating from the graph, if $f(3)=16.8$, then the average rate of change $\frac{f(3)-f(2)}{1}=\frac{16.8 - 9.60}{1}=7.2$.

Step5: When $h = 0.4$

The point $(2 + h,f(2 + h))=(2.4,f(2.4))$. Estimating from the graph, if $f(2.4)=12.48$, then the average rate of change $\frac{f(2.4)-f(2)}{0.4}=\frac{12.48 - 9.60}{0.4}=\frac{2.88}{0.4}=7.2$.

Step6: When $h = 0.1$

The point $(2 + h,f(2 + h))=(2.1,f(2.1))$. Estimating from the graph, if $f(2.1)=10.32$, then the average rate of change $\frac{f(2.1)-f(2)}{0.1}=\frac{10.32 - 9.60}{0.1}=\frac{0.72}{0.1}=7.2$.

Step7: For part b

As $h$ gets smaller and smaller, the difference quotient $\frac{f(2 + h)-f(2)}{h}$ approaches the instantaneous rate of change of the function $f$ at $x = 2$. So the statement "As the value of $h$ gets smaller and smaller, the value of the difference quotient becomes a better and better approximation of the average rate of change of a function $f$ near the point $(2,f(2))$" is True.

Answer:

i. $7.68$
ii. $7.2$
iii. $7.2$
iv. $7.2$
v. $7.2$
b. True