Chapter 14, Problem 42RE

To determine

To find: $f\left(1\right)$ and $f\left(4\right)$ .

Answer to Problem 42RE

  $f\left(1\right)=1$ and $f\left(4\right)=\frac{1}{16}$ ;

Explanation of Solution

Given information:

  $f\left(x\right)=\frac{1}{x^2},\left[1,4\right]$

Solution:

  $\begin{array}{l}f\left(1\right)=\frac{1}{1^2}\\ \Rightarrow f\left(1\right)=1\end{array}$

  $\begin{array}{l}f\left(4\right)=\frac{1}{4^2}\\ \Rightarrow f\left(4\right)=\frac{1}{16}\end{array}$

Domain and Range of $f\left(x\right)=\frac{1}{x^2}$ ;

Domain: $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Range: $\left(0,\infty \right)$

Hence, $f\left(1\right)=1$ and $f\left(4\right)=\frac{1}{16}$