Question

The function f is defined above. For what value of k , if any, is f continuous at x=2?

Answer 1

Answer: 3

Given:

$f(x)=\begin{cases}x^2-3x+9{\text{ }for\text{ }x\leq2} \\ kx+1\text{ }{\text{ }for\text{ }x>2}\end{cases}$

First, we find the limit of the function approaching from both sides when x=2:

Left-hand limit:

$\begin{gathered} \lim_{x\to2^-}x^2-3x+9 \\ =(2)^2-3(2)+9 \\ =4-6+9 \\ =7 \end{gathered}$

Right-hand limit:

$\begin{gathered} \lim_{x\to2^+}kx+1 \\ =k(2)+1 \\ =2k+1 \end{gathered}$

Hence,

$\begin{gathered} 2k+1=7 \\ 2k=7-1 \\ 2k=6 \\ k=\frac{6}{2} \\ k=3 \end{gathered}$

Therefore, the answer would be k=3