Chapter 2, Problem 117CR

a

To determine

To find : the domain of f.

Answer to Problem 117CR

  $R-\left\{-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}\right\}$

Explanation of Solution

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

Calculation:

Domain of a function f(x) is set of values of x for which the function f(x) is defined.

The given function f(x) is defined for all real values of x for which the denominator of the function f(x) is non-zero.

By setting denominator of function f(x) equal to 0, we get

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

Thus, domain of given function is set of all real numbers except $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$ , that is domain of the function is $R-\left\{-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}\right\}$

b

To determine

To decide : whether f is continuous..

Answer to Problem 117CR

Not continuous

Explanation of Solution

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

Calculation:

Since f is defined at the points $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$ , therefore f is not continuous at these points.

Therefore, function f is not continuous.

c

To determine

To identify : any horizontal and vertical asymptotes.

Answer to Problem 117CR

Vertical asymptotes at $x=-\sqrt{\frac{3}{2}}$ and $x=\sqrt{\frac{3}{2}}$

Horizontal asymptotes at $y=2$

Explanation of Solution

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

Calculation:

The vertical asymptotes occur at those values of x for which denominator is equal to zero.

Since denominator is zero for $x=\pm \sqrt{\frac{3}{2}}$

Therefore, graph of given function have vertical asymptotes at $x=-\sqrt{\frac{3}{2}}$ and $x=\sqrt{\frac{3}{2}}$ .

To find the horizontal asymptote, we note that the degree of the numerator is 2 and degree of the denominator is also 2. Since the degrees of both numerator and denominator is same, therefore, the graph will have a horizontal asymptote at $y=\frac{4}{2}$ , that is at $y=2$