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=(6)/(2) \\ k=3 \end{gathered}`

Therefore, the answer would be k=3