Chapter 2, Problem 131RE

a.

To determine

To find the domain of the function $f(x)=\frac{4x^2}{2x^2-3}$ .

Answer to Problem 131RE

Domain: $\left(-\infty ,-\frac{\sqrt{3}}{2}\right)\cup \left(-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right)\cup \left(\frac{\sqrt{3}}{2},\infty \right)$

Explanation of Solution

Given information:

Function: $f(x)=\frac{4x^2}{2x^2-3}$

Formula used:

To find the zeros of the quadratic equation

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

To find the domain of a function,

Check for the values of $x$ which leaves the denominator 0.

For $2x^2-3$ being the denominator of the function,

Find the zeros of the function as follows:

  $2x^2-3=0$

Applying the quadratic formula,

  $\begin{array}{l}x=\frac{0\pm \sqrt{{(0)}^2-4(2)(-3)}}{2(2)}\\ x=\frac{\pm \sqrt{24}}{4}\\ x=\frac{\pm 2\sqrt{3}}{4}\\ x=\frac{\pm \sqrt{3}}{2}\end{array}$

At $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$ the denominator becomes 0.

So, the domain of the given function is all the real numbers except for $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$ .

Conclusion:

Therefore, the domain of $f(x)=\frac{4x^2}{2x^2-3}$ is $\left(-\infty ,-\frac{\sqrt{3}}{2}\right)\cup \left(-\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2}\right)\cup \left(\frac{\sqrt{3}}{2},\infty \right)$ .

b.

To determine

To check whether the function $f(x)=\frac{4x^2}{2x^2-3}$ is continuous or not

Answer to Problem 131RE

Not continuous at $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

Explanation of Solution

Given information:

Function: $f(x)=\frac{4x^2}{2x^2-3}$

In subpart (a), it is observed that the function becomes infinity if $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$ .

Therefore, the function is not continuous at $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$ .

Conclusion:

Therefore, the function is not continuous at $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$ .

c.

To determine

To find the vertical and horizontal asymptotes of the function $f(x)=\frac{4x^2}{2x^2-3}$

Answer to Problem 131RE

Vertical asymptote: $x=\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}$

Horizontal asymptote: $y=2$

Explanation of Solution

Given information:

Function: $f(x)=\frac{4x^2}{2x^2-3}$

Calculation:

Vertical asymptotes:

To find vertical asymptotes, put denominator of the given function equal to zero.

  $\Rightarrow 2x^2-3=0$

Applying quadratic formula:

  $\begin{array}{l}x=\frac{0\pm \sqrt{{(0)}^2-4(2)(-3)}}{2(2)}\\ x=\frac{\pm \sqrt{24}}{4}\\ x=\frac{\pm 2\sqrt{3}}{4}\\ x=\frac{\pm \sqrt{3}}{2}\end{array}$

Horizontal asymptotes:

As the degree of numerator is equal to the degree of the denominator, consider the variable with the highest degree in both numerator and denominator.

The horizontal asymptote is:

  $\begin{array}{l}y=\frac{4x^2}{2x^2}\\ y=2\end{array}$