Chapter 9.3, Problem 5QR

To determine

To identify: The smallest number M that bounds $\left|f\right|$ from the above on the interval I such that $\left|f\left(x\right)\le M\right|$ for all x in I .

Answer to Problem 5QR

The value of M is 7.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\left\{\begin{array}{ll}2-x^2,\hfill & x\le 1,\hfill \\ 2x-1,\hfill & x>1,\hfill \end{array}I=[-3,3]\right.$ .

Consider the given function.

In the interval $\left[-3,1\right],f\left(x\right)=2-x^2$ .

The function has local maximum or local minimum for x where first derivative is zero.

  $\begin{array}{c}f\text{'}\left(x\right)=-2x\\ 0=-2x\\ x=0\end{array}$

The critical point is in the interval $\left[-3,1\right]$ .

Check the value at $x=0$ .

  $\left|f\left(0\right)\right|=\left|2\right|\mathrm{...}\left(1\right)$

At $x=-3$

  $\begin{array}{c}\left|f\left(x\right)\right|=\left|2-x^2\right|\\ \left|f\left(-3\right)\right|=\left|2-9\right|\\ \text{=7}\end{array}$

At $x=1$

  $\begin{array}{c}\left|f\left(x\right)\right|=\left|2-x^2\right|\\ \left|f\left(1\right)\right|=\left|2-1\right|\\ \text{=1 }\mathrm{...}\left(2\right)\end{array}$

In the interval $\left(1,3\right],f\left(x\right)=2x-1$ .

The function $f\left(x\right)$ has local maxima or local minima for x where first derivative is zero.

  $f\text{'}\left(x\right)=2$

Therefore, first derivative is always positive, thus the function is strictly increasing for $\left(1,3\right]$ and now check the value at the end point.

At $x=3$

  $\begin{array}{c}\left|f\left(x\right)\right|=\left|2x-1\right|\\ \left|f\left(3\right)\right|=\left|6-1\right|\\ \text{=5 }\mathrm{...}\left(3\right)\end{array}$

From equation (1), (2) and (3) it can be observed that $\left|f\left(x\right)\right|\le \text{7 on }\left[-3,3\right]$ .

Thus, the value of M should be 7.