Q. 27 - Precalculus Enhanced with Graphing Utilities

Question

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

(a) Is the point $\left(-1, 1\right)$ on the graph of $f$?

(b) If $x=2$, what is $f\left(x\right)$? What point is on the graph of $f$?

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

(d) What is the domain of $f$?

(e) List the x-intercepts, if any, of the graph of $f$.

(f) List the y-intercept, if there is one, of the graph of $f$. 

Step-by-Step Solution

Answer

Part (a) Yes

Part (b) $f\left(2\right)=\frac{8}{17}$; The point on the graph is $\left(2,\frac{8}{17}\right)$.

Part (c) Values of $x$ are $-1 and 1$; Points on the graph are $\left(-1, 1\right) and \left(1, 1\right)$.

Part (d) The domain is all real numbers.

Part (e) The x-intercept is 0.

Part (f) The y-intercept is 0.

Part (a) Step 1. Find the value of the function given point.

When $x=-1$,

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

The point $\left(-1, 1\right)$ is on the graph of $f$.

Part (b) Step 1. Find the value at 2.

If $x=2$, then

$\begin{array}{rcl}f\left(2\right)& =& \frac{2{\left(2\right)}^2}{{\left(2\right)}^4+1}\\ & =& \frac{8}{16+1}\\ & =& \frac{8}{17}\end{array}$

The point $\left(2,\frac{8}{17}\right)$ is on the graph.

Part (c) Step 1. Find the value of x .

If $f\left(x\right)=1$, then

$\begin{array}{rcl}f\left(x\right)& =& 1\\ \frac{2x^2}{x^4+1}& =& 1\\ 2x^2& =& x^4+1\\ x^4-2x^2+1& =& 0\\ {\left(x^2-1\right)}^2& =& 0\\ x^2-1& =& 0\\ \left(x+1\right)\left(x-1\right)& =& 0\\ x+1& =& 0 or x-1=0\\ x& =& -1 or x=1\end{array}$

The point on the graph are $\left(-1, 1\right) and \left(1, 1\right)$.

Part (d) Step 1. Find the domain.

The denominator of function $f$ will never be zero for any real value of $x$. So, the domain of $f$ is the set of all real numbers. 

Part (e) Step 1. Find x -intercepts.

The x-intercepts of the graph of $f$ are the real solutions of the equation $f\left(x\right)=0$ that are in the domain of $f$.

The real solution of the equation$f\left(x\right)=\begin{array}{rcl}& & \frac{2x^2}{x^4+1}\end{array}$ is

$\begin{array}{rcl}f\left(x\right)& =& 0\\ \frac{2x^2}{x^4+1}& =& 0\\ 2x^2& =& 0\\ x^2& =& 0\\ x& =& 0\end{array}$

 The x-intercept is $0$.

Part (f) Step 1. Find y -intercepts.

The y-intercepts of the graph of $f$ are the real values of $f\left(0\right)$.

$\begin{array}{rcl}f\left(x\right)& =& \frac{2x^2}{x^4+1}\\ f\left(0\right)& =& \frac{2{\left(0\right)}^2}{{\left(0\right)}^4+1}\\ & =& 0\end{array}$

The y-intercept is $0$.