Q. 12 - Calculus

Question

Consider the function f graphed here: 

(a) Sketch secant lines from (2, f(2)) to (2 + h, f(2 + h)) on the graph of f , for the following values of h: h = 0.5, h = 0.25, h = 0.1, h = −0.5, h = −0.25, and h = −0.1. 

(b) Use the secant lines you sketched in part (a) to graphically evaluate $\underset{h\to 0^{+}}{lim} \frac{f(2+h)-f\left(2\right)}{h}$ and lim $\underset{h\to 0^{-}}{lim} \frac{f(2+h)-f\left(2\right)}{h}$.

(c) Use your answer from part (b) to show that f is not differentiable at x = 2. 

Step-by-Step Solution

Answer

Part (a) Step 1. Given information.

We have been given a graph for the function f:

We have to sketch secant lines from $(2, f(2\left)\right) to (2 + h, f(2 + h\left)\right)$ on the graph of f , for the following values of $h: h = 0.5, h = 0.25, h = 0.1, h = -0.5, h = -0.25, \mathrm{and} h = -0.1.$

Part (a) Step 2. Sketching secant lines

From the graph, it is observed that the value of the function at $x=2$ is 2. For different values of h. the obtained points are $(2.5,2), (2.25,2), (2.1,2), (1.5,2), (1.75,1.5) \mathrm{and} (1.9,1.4).$

The first secant line passes through the point (2,2) and (2.5,2); the second secant line passes through the line (2,2) and (2.25,2); the third secant line passes through the line (2,2) and (2.1,2); the fourth secant line passes through the line (2,2) and (1.5,2.5); fifth secant line passes through the line (2,2) and (1.75,1.85); and the sixth line passes through the line (2,2) and (1.9,1.4).

Join the points through straight lines to graph each secant line on the given graph.

Part (b) Step 1. Evaluating the function

Recall that for a function $f,\frac{f(c+h)-f\left(c\right)}{h}$ denotes the slope of the different secant lines around the point x=c where c is a real number lying in the domain of  f.

As $h\to 0^{+},\frac{f(c+h)-f\left(c\right)}{h}$ denotes the value that the slope of the secant lines approaches from the right.

As $h\to 0^{-},\frac{f(c+h)-f\left(c\right)}{h}$ denotes the value that the slope of the secant lines approaches from the left.

From the graph,

the secant lines become vertical as the value approaches 0 from the right side,

This implies that the value of the limit $\underset{h\to 0^{+}}{lim} \frac{f(2+h)-f\left(2\right)}{h}$ is 0

The secant lines become vertical as the value approaches 0 from the left side,

Since the slope approaches $\infty$ as $h\to 0^{-}$

Thus,

$\underset{h\to 0^{-}}{lim} \frac{f(2+h)-f\left(2\right)}{h}$ does not exist.

Part (c) Step 1. Show that f is not differentiable at x = 2.

A function f is said to be a differentiable function when the left-hand derivative and the right-hand derivative of the function exist and are equal. From part (b), it is calculated that the left-hand derivative does not exist, whereas the right-hand derivative is 0.

Therefore, the function is not differentiable at x = 2.