Q. 87 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\frac{2x-A}{x-3}$ and $f\left(4\right)=0$, what is the value of $A$? Where is $f$ not defined?

Step-by-Step Solution

Answer

The value of $A=8$ and at $x=3$ function is not defined.

Step 1. Given Information

If $f\left(x\right)=\frac{2x-A}{x-3}$ and $f\left(4\right)=0$

We have to find the value of $A$ and where is $f$ not defined.

Step 2. To find the value of A we firstly find the f ( 4 ) by putting x = 4 in the given function.

$$\begin{gathered} f\left(4\right)=\frac{2\times 4-A}{4-3} \\ f\left(4\right)=\frac{8-A}{1} \\ f\left(4\right)=8-A \end{gathered}$$

Step 3. Now the function is f ( 4 ) = 8 - A . We know f ( 4 ) = 0 so we put the value of f ( 4 ) in the function.

$0=8-A$

Subtract 8 on both side

$$\begin{gathered} 0-8=8-A-8 \\ -8=-A \\ -A=-8 \end{gathered}$$

Multiply with $-1$ on both side

$$\begin{gathered} -A\times (-1)=-8\times (-1) \\ A=8 \end{gathered}$$

Step 4. To find, where is f is not defined, we put denominator of function equal to zero.

$x-3=0$

Add 3 on both side

$$\begin{gathered} x-3+3=0+3 \\ x=3 \end{gathered}$$

At $x=3$ the function is not defined.