Q. 88 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\frac{x-B}{x-A}$, $f\left(2\right)=0$ and $f\left(1\right)$ is undefined, what is the value of $A \mathrm{and} B$?

Step-by-Step Solution

Answer

The value of $B=-2$ and the value of $A=-1$.

Step 1. Given Information

If $f\left(x\right)=\frac{x-B}{x-A}$, $f\left(2\right)=0$ and $f\left(1\right)$ is undefined,

We have to find the value of $A \mathrm{and} B$.

Step 2. To find the value of B we firstly find the f ( 2 ) by putting x = 2 in the given function.

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

Step 3. Now the function is f ( 2 ) = 2 - B 2 - A . We know f ( 2 ) = 0 so we put the value of f ( 2 ) in the function.

$0=\frac{2-B}{2-A}$

Multiply with $2-A$ on both side

$$\begin{gathered} 0\times (2-A)=\frac{2-B}{2-A}\times (2-A) \\ 0=2-B \\ 2-B=0 \end{gathered}$$

Add 2 on both side

$$\begin{gathered} 2-B+2=0+2 \\ -B=2 \end{gathered}$$

Multiply with $-1$ on both side

$$\begin{gathered} -B\times (-1)=2\times (-1) \\ B=-2 \end{gathered}$$

Step 4. To find the value of A we firstly find f ( 1 ) by putting x = 1 on the given function.

$f\left(1\right)=\frac{1-B}{1-A}$

Since $f\left(1\right)$ is not defined so

$1-A=0$

Add 1 on both side

$$\begin{gathered} 1-A+1=0+1 \\ -A=1 \end{gathered}$$

Multiply with $-1$ on both side

$$\begin{gathered} -A\times (-1)=1\times (-1) \\ A=-1 \end{gathered}$$