Q. 28 - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=\frac{2x}{x-2}$

(a) Is the point $\left(\frac{1}{2},-\frac{2}{3}\right)$ on the graph of $f$?

(b) If $x=4$, 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) $f\left(4\right)=4$; The point on the graph is $\left(4, 4\right)$.

Part (c) Value of $x$ is $-2$; Point on the graph is $\left(-2, 1\right)$.

Part (d) The domain is $\left\{x\overline{)x\ne }2\right\}$.

Part (e) The x-intercept is 0.

Part (f) The y-intercept is 0.

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

When $x=\frac{1}{2}$, Substitute in the given function,

 

$\begin{array}{rcl}f\left(x\right)& =& \frac{2x}{x-2}\\ f\left(\frac{1}{2}\right)& =& \frac{2\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)-2}\\ & =& \frac{1}{-{\displaystyle \frac{3}{2}}}\\ & =& -\frac{2}{3}\end{array}$

The point $\left(\frac{1}{2},-\frac{2}{3}\right)$ is on the graph of $f$.

Part (b) Step 1. Find the value at 4.

If $x=4$,substitute this value in the given function.

$\begin{array}{rcl}f\left(4\right)& =& \frac{2\left(4\right)}{4-2}\\ & =& \frac{8}{2}\\ & =& 4\end{array}$

The point $\left(4, 4\right)$ is on the graph $f$.

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

If $f\left(x\right)=1$, then find the value of x.

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

The point on the graph is $\left(-2, 1\right)$.

Part (d) Step 1. Find the domain.

The domain of $f$ is $\left\{x\overline{)x\ne }2\right\}$, since $x=2$ results in division by 0.

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

$f\left(x\right)=\frac{2x}{x-2}$is

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

 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 $f\left(0\right)$.

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

The y-intercept is $0$.