Q. 11 - Calculus

Question

For the graph of f appearing next at the left, label each of the following quantities to illustrate that 

$f^{\mathrm{\prime }}\left(c\right)\approx \frac{f(c+h)-f\left(c\right)}{h}$

(a) the locations c, c + h, f(c), and f(c + h) 

(b) the distances h and f(c + h) − f(c) 

(c) the slopes $\frac{f(c+h)-f\left(c\right)}{h}$ and $f^{\mathrm{\prime }}\left(c\right)$

Step-by-Step Solution

Answer

(a) 

(b) h=1 and f(c+h)-f(c)=0.5

(c) Slopes is 0.5 and $f^{\text{'}}\left(c\right)=0.5$

Part (a) Step 1. Given information.

For the graph of f appearing next at the left, label each of the following quantities to illustrate that

$f^{\mathrm{\prime }}\left(c\right)\approx \frac{f(c+h)-f\left(c\right)}{h}$

Part (a) Step 2. Locate c , c + h , f ( c ) , f ( c + h )

After locating the point, the graph is :

Part (b) Step 1. Find the distance h and f(c+h)-f(c)

By looking at the graph, we can said that the values of h is 1 unit in the graph and $f(c+h)-f\left(c\right)$ is 0.5

Part (c) Step 1. Find the slope and f ' ( c )

We have $f(c+h)-f\left(c\right)=0.5$ and h=1, therefore the slopes is:

$\begin{array}{r}\frac{f(c+h)-f\left(c\right)}{h}=\frac{0.5}{1}\\ =0.5\end{array}$

Again,

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(c\right)=\underset{h\to 0}{lim} \frac{f(c+h)-f\left(c\right)}{h}\\ =\underset{h\to 0}{lim} \frac{0.5}{1}\\ =0.5\end{array}$