Question

Determine the value of pif f (x) is continuous at x = 3. Refer to the picture given.

Answer 1

Given:-

$f(x)=\begin{cases}\frac{x^2-12x+27}{x-3} \\ 6px\end{cases}$

To find the value of x=3.

From the question it is clear that the value of 3 lies within,

$\frac{x^2-12x+27}{x-3}$

Now we substituting the value of x=3. we get,

$\begin{gathered} f(x)=\frac{x^2-12x+27}{x-3}_{} \\ f(3)=\frac{3^2-12\times3+27}{3-3} \\ f(3)=\frac{9-36+27}{0} \\ f(3)=\frac{0}{0} \end{gathered}$

So the value of f(3)=0