Chapter 5.2, Problem 47E

To determine

To show : $f\left(x\right)=0$ exactly once between $a$ and $b$ .

Answer to Problem 47E

The possible explanation is given below.

Explanation of Solution

Given information :

Assume that $f$ is continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ . Also assume that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs and $f\text{'}\ne 0$ between $a$ and $b$ .

Calculation :

Since $f\text{'}\ne 0$ and $f\left(a\right)$ and $f\left(b\right)$ have opposite signs. Therefore, $f$ is always increasing or decreasing on $\left[a,b\right]$ . So there is no local extrema.

Now, as $f$ is always increasing and continuous this show that $f\left(x\right)=0$ exactly once between $a$ and $b$ .