Chapter 2, Problem 4RE

a.

To determine

To find the value of $f\left(2\right)$ .

Answer to Problem 4RE

  $f\left(2\right)=\frac{4}{3}$

Explanation of Solution

Given:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Put $x=2$ in $f\left(x\right)=\frac{x^2}{x+1}$ , we get

  $\begin{array}{c}f\left(2\right)=\frac{2^2}{2+1}\\ =\frac{4}{3}\end{array}$

b.

To determine

To find the value of $f\left(-2\right)$ .

Answer to Problem 4RE

  $f\left(-2\right)=-4$

Explanation of Solution

Given information:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Put $x=-2$ in $f\left(x\right)=\frac{x^2}{x+1}$ , we get

  $\begin{array}{c}f\left(-2\right)=\frac{{\left(-2\right)}^2}{-2+1}\\ =\frac{4}{-1}\\ =-4\end{array}$

c.

To determine

To find the value of $f\left(-x\right)$ .

Answer to Problem 4RE

  $f\left(-x\right)=\frac{x^2}{1-x}$

Explanation of Solution

Given information:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Put $x=-x$ in $f\left(x\right)=\frac{x^2}{x+1}$ , we get

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

d.

To determine

To find the value of $-f\left(x\right)$ .

Answer to Problem 4RE

  $-f\left(x\right)=-\frac{x^2}{x+1}$

Explanation of Solution

Given information:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Multiply minus sign to both sides

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

e.

To determine

To find the value of $f\left(x-2\right)$ .

Answer to Problem 4RE

  $f\left(x-2\right)=\frac{x^2-4x+4}{x-1}$

Explanation of Solution

Given information:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Put $x=x-2$ in $f\left(x\right)=\frac{x^2}{x+1}$ , we get

  $\begin{array}{c}f\left(x-2\right)=\frac{{\left(x-2\right)}^2}{x-2+1}\\ =\frac{x^2-4x+4}{x-1}\end{array}$

f.

To determine

To find the value of $f\left(2x\right)$ .

Answer to Problem 4RE

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

Explanation of Solution

Given information:

A function is given as

  $f\left(x\right)=\frac{x^2}{x+1}$

Calculation:

Put $x=2x$ in $f\left(x\right)=\frac{x^2}{x+1}$ , we get

  $\begin{array}{c}f\left(2x\right)=\frac{{\left(2x\right)}^2}{2x+1}\\ =\frac{4x^2}{2x+1}\\ \end{array}$