Chapter 6, Problem 1RE

If $f\left(x\right)=\frac{x^2+2x-3}{x^2-4}$, find the following function values. If a function value does not exist, so state.

$f\left(4\right)$

$f\left(0\right)$

$f\left(-2\right)$

$f\left(-3\right)$

To determine

To calculate: The Function values at $f\left(4\right)$.

Answer to Problem 1RE

Solution:

The required value is $f\left(4\right)=\frac{7}{4}$.

Explanation of Solution

Given:

The function is:

$f\left(x\right)=\frac{x^2+2x-3}{x^2-4}$.

Calculation:

To find $f\left(4\right)$, replace x with 4.

$\begin{array}{c}f\left(4\right)=\frac{{\left(4\right)}^2+2\left(4\right)-3}{{\left(4\right)}^2-4}\\ =\frac{16+8-3}{16-4}\\ =\frac{21}{12}\\ =\frac{7}{4}\end{array}$

Hence, the value is $f\left(4\right)=\frac{7}{4}$.

To determine

To calculate: The function values at $f\left(0\right)$.

Answer to Problem 1RE

Solution:

The required value is $f\left(0\right)=\frac{3}{4}$.

Explanation of Solution

Given:

The function is:

$f\left(x\right)=\frac{x^2+2x-3}{x^2-4}$

Calculation:

To find $f\left(0\right)$, replace x with 0.

$\begin{array}{c}f\left(0\right)=\frac{{\left(0\right)}^2+2\left(0\right)-3}{{\left(0\right)}^2-4}\\ =\frac{-3}{-4}\\ =\frac{3}{4}\end{array}$

Hence, the value is $f\left(0\right)=\frac{3}{4}$.

To determine

To calculate: The function values at $f\left(-2\right)$.

Answer to Problem 1RE

Solution:

$f\left(-2\right)$ does not exist.

Explanation of Solution

Given:

The function is:

$f\left(x\right)=\frac{x^2+2x-3}{x^2-4}$.

To determine

To calculate: The function values at $f\left(-3\right)$.

Answer to Problem 1RE

Solution:

The required value is $f\left(-3\right)=0$.

Explanation of Solution

Given:

The function is:

$f\left(x\right)=\frac{x^2+2x-3}{x^2-4}$.

Calculation:

To find $f\left(-3\right)$, replace x with $-3$.

$\begin{array}{c}f\left(4\right)=\frac{{\left(-3\right)}^2+2\left(-3\right)-3}{{\left(-3\right)}^2-4}\\ =\frac{9-6-3}{9-4}\\ =\frac{0}{-5}\\ =0\end{array}$

Hence, the value is $f\left(-3\right)=0$.