Chapter 2.2, Problem 16E

To determine

To sketch: The graph of $f(x)$ which satisfies the conditions, $\underset{x\to 0^{-}}{\lim }f(x)=2,\underset{x\to 0^{+}}{\lim }f(x)=0,$ $\underset{x\to 4^{-}}{\lim }f(x)=3,\underset{x\to 4^{+}}{\lim }f(x)=0,f\left(0\right)=2\text{ and }f\left(4\right)=1$.

Explanation of Solution

Graph:

Given that, $\underset{x\to 0^{-}}{\lim }f(x)=2,\underset{x\to 0^{+}}{\lim }f(x)=0,\underset{x\to 4^{-}}{\lim }f(x)=3,\underset{x\to 4^{+}}{\lim }f(x)=0,f\left(0\right)=2\text{ and }f\left(4\right)=1$.

Here, the left hand limit and the right hand limit of $\underset{x\to 0}{\lim }f(x)\text{ and }\underset{x\to 4}{\lim }f(x)$ are not the same.

Hence, the limits, $\underset{x\to 0}{\lim }f(x)\text{ and }\underset{x\to 4}{\lim }f(x)$ does not exist.

Moreover, according to the given conditions the graph must contain two points such that (0, 2) as $f\left(0\right)=2$ and (4, 1) as $f\left(4\right)=1$.

Thus, the possible graph of $f(x)$ is shown below in Figure 1.

In Figure 1, open dots indicate that the points are excluded.