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?

Short answer

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

Step-by-step solution

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.