Chapter 2.4, Problem 38E

To determine

To sketch:two possible graphsof ffirst graph shows that $f\left(x\right)$ is not satisfy the conclusion of the IVT and other shows that $f\left(x\right)$ issatisfy the conclusion ofthe IVT.

Explanation of Solution

Given:

  $f\left(x\right)$ is continuous on $\left[0,1\right]$ except at 0.25 and that $f\left(0\right)=1,f\left(1\right)=3$ .

Let $N=2$

Calculation:

Consider both the conditions:

In the first graph $f\left(x\right)$ does not take on every value, including $N=2$ , between $y=1,\text{ and }y=3$

The graph is given below:

In the second graph $f\left(x\right)$ does not take on every value, between $y=1,\text{ and }y=3$ , including $N=2$ twice, despite there being a discontinuity.

The graph is given below:

Hence, a function $f\left(x\right)$ could take on every value between $f\left(a\right)\text{ and }f\left(b\right)$ even if the hypothesis of the I

VT are not satisfied.