Chapter 2.2, Problem 27AYU

In problems $25-30$, answer the questions about each function.

$f(x)=\frac{12x^4}{x^2+1}$

Is the point $(-1,6)$ on the graph of $f$?

If $x=3$, what is $f(x)$? What point is on the graph of $f?$

If $f(x)=1,$ what is x? What point(s) are on the graph of $f?$

What is the domain of $f?$

List the $x-$ intercepts, if any, of the graph of $f?$

List the $y-$ intercept, if there is one, of the graph of $f?$

To determine

The point $\left(-1,6\right)$ on the graph of $f$ if $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

Yes, the point $\left(-1,6\right)$ is on the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Explanation of Solution

Given Information:

$f\left(x\right)=\frac{12x^4}{x^2+1}$, and the point $\left(-1,6\right)$.

Explanation:

If $\left(x,y\right)$ is a point on the graph of a function $f$, then $y$ is the value of $f$ at $x$, that is, $y=f(x)$.

Substitute $x=-1$ in $f\left(x\right)=\frac{12x^4}{x^2+1}$.

$f\left(-1\right)=\frac{12{\left(-1\right)}^4}{{\left(-1\right)}^2+1}$,

$f\left(-1\right)=\frac{12}{2}$,

$\Rightarrow f\left(-1\right)=6$.

Thus, the point $\left(-1,6\right)$ is on the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

To determine

The value of $f\left(x\right)$ if $x=3$. Find the point on the graph of $f$ if $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

If $x=3$, the value of $f\left(x\right)=\frac{486}{5}$, and the point $\left(3,\frac{486}{5}\right)$ is on the graph of $f$.

Explanation of Solution

Given Information:

The function $f\left(x\right)=\frac{12x^4}{x^2+1}$ if $x=3$.

Explanation:

If $x=3$ then,

$f\left(3\right)=\frac{12{\left(3\right)}^4}{3^2+1}$,

$f\left(3\right)=\frac{\left(12\right)\left(81\right)}{9+1}$,

$\Rightarrow f\left(3\right)=\frac{972}{10}=\frac{486}{5}$.

Thus, $f\left(x\right)=\frac{486}{5}$ if $x=3$.

The point $\left(3,\frac{486}{5}\right)$ is on the graph $f$.

To determine

The value of $x$ if $f\left(x\right)=1$. Find the points on the graph of $f$ if $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

The values of $x=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$ if $f\left(x\right)=1$. The points $\left(-\frac{1}{\sqrt{3}},1\right),\left(\frac{1}{\sqrt{3}},1\right)$ are on the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Explanation of Solution

Given Information:

The functions $f\left(x\right)=\frac{12x^4}{x^2+1}$ and $f\left(x\right)=1$.

Explanation:

Substitute $f\left(x\right)=1$ in $f\left(x\right)=\frac{12x^4}{x^2+1}$.

$\Rightarrow 1=\frac{12x^4}{x^2+1}$,

$\Rightarrow 12x^4=x^2+1$,

$12x^4-x^2-1=0$.

By using substitution $u=x^2$, above equation would be,

$12u^2-u-1=0$.

Find factors of the quadratic equation,

$12u^2-4u+3u-1=0$,

$4u\left(3u-1\right)+1\left(3u-1\right)=0$,

$\left(3u-1\right)\left(4u+1\right)=0$,

$3u-1=0\Rightarrow u=\frac{1}{3}$.

$4u+1=0\Rightarrow u=-\frac{1}{4}$.

Now resubstitute values of $u$,

$x^2=\frac{1}{3},x^2=-\frac{1}{4}$.

But square of the term cannot be negative, so $x^2=-\frac{1}{4}$ is not possible.

$x^2=\frac{1}{3}$.

Taking square root on both sides,

$x=\pm \frac{1}{\sqrt{3}}$.

$x=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.

If $f\left(x\right)=1$, then $x=-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.

The points $\left(-\frac{1}{\sqrt{3}},1\right),\left(\frac{1}{\sqrt{3}},1\right)$ are on the graph of $f\left(x\right)=3x^2+x-2$.

Thus, $f\left(x\right)=-2$, then $x=0$ or $x=-\frac{1}{3}$. The points $\left(0,-2\right)$ and $\left(-\frac{1}{3},-2\right)$ are on the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

To determine

The domain of $f$ if $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

The domain of $f\left(x\right)=\frac{12x^4}{x^2+1}$ is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given Information:

$f\left(x\right)=\frac{12x^4}{x^2+1}$.

Explanation:

$f\left(x\right)=\frac{12x^4}{x^2+1}$ is a rational function.

The domain of the function is set of all values of $x$ for which the function is defined.

Since it is a rational function, it is defined for all values of $x$ except when denominator is 0.

Set denominator of the above function to 0,

$x^2+1=0\Rightarrow x^2=-1$

But square of the term cannot be negative. So this is not possible.

Thus, domain of the function $f\left(x\right)=\frac{12x^4}{x^2+1}$ is all real numbers. In interval notation, the domain would be $\left(-\infty ,\infty \right)$.

To determine

The $x$-intercepts, if any, of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

The $x$-intercept of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$ is $\left(0,0\right)$.

Explanation of Solution

Given Information:

The function is $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Explanation:

The $x$-intercepts of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$ are the real solutions of the equation $f\left(x\right)=0$ that are in the domain of $f$.

$f\left(x\right)=\frac{12x^4}{x^2+1}=0$.

$12x^4=0$,

$\Rightarrow x=0$.

Thus, the $x$-intercept of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$ is $\left(0,0\right)$.

To determine

The $y$-intercepts, if any, of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$.

Answer to Problem 27AYU

Solution:

The $y$-intercept of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$ is $\left(0,0\right)$.

Explanation of Solution

Given Information:

$f\left(x\right)=\frac{12x^4}{x^2+1}$.

Explanation:

To find the $y$-intercepts of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$, substitute $x=0$ in $f\left(x\right)$.

$f\left(x\right)=\frac{12x^4}{x^2+1}$.

Substitute $x=0$ in above function,

$f\left(0\right)=\frac{12{\left(0\right)}^4}{0^2+1}=\frac{0}{1}$,

$\Rightarrow f\left(0\right)=0$.

Thus, the $y$-intercept of the graph of $f\left(x\right)=\frac{12x^4}{x^2+1}$ is $\left(0,0\right)$.