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}(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,

`(x^2-12x+27)/(x-3)`

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

`\begin{gathered} f(x)=(x^2-12x+27)/(x-3)_{} \\ f(3)=(3^2-12*3+27)/(3-3) \\ f(3)=(9-36+27)/(0) \\ f(3)=(0)/(0) \end{gathered}`

So the value of f(3)=0