Chapter 2.8, Problem 41STP

To determine

Identify function for range $\left|f\left(x\right)\right|f\left(x\right)\le 0$ .

Answer to Problem 41STP

  $\left(J\right)f\left(x\right)=-\left|x\right|$

Explanation of Solution

Given information:

The functions,

  $\begin{array}{l}Ff\left(x\right)=-xHf\left(x\right)=\left|x\right|\\ Gf\left(x\right)=\left[x\right]Jf\left(x\right)=-\left|x\right|\end{array}$

Calculation:

Consider the given range, $\left|f\left(x\right)\right|f\left(x\right)\le 0$ that is for all negative and zero value of function.

First consider $\left(F\right).f\left(x\right)=-x$ function,

Substitute,

  $\begin{array}{l}x=1inf\left(x\right)=-x\\ f\left(x\right)=-1<0\end{array}$

So this is not the function.

Now consider $\left(H\right)f\left(x\right)=\left|x\right|$ ,

For any value of $x$ the range of the function is always greater than zero

The range of $f\left(x\right)=\left|x\right|is\left|f\left(x\right)\right|f\left(x\right)>0$

Hence, this is not the function.

Now consider $\left(G\right)f\left(x\right)=\left[x\right]$ ,

Substitute,

  $\begin{array}{l}x=1inf\left(x\right)=\left[x\right]\\ f\left(x\right)=1>0\end{array}$

This is not the range of the function.

Now consider, $\left(J\right)f\left(x\right)=-\left|x\right|$

For any value of $x$ the range of the function is always less than or equal to zero.

The range of function $f\left(x\right)=-\left|x\right|is\left|f\left(x\right)\right|f\left(x\right)<0ifx\ne 0and\left|f\left(x\right)\right|f\left(x\right)<0ifx=0$

Hence, $f\left(x\right)=-\left|x\right|$ is the function whose range is $\left|f\left(x\right)\right|f\left(x\right)\le 0$