For what valve of k is the following function continuous at x = 2 ƒ(x) = {2x + 1, when x < 2 k, when x = 2 3x - 1,when x > 2

Question

For what valve of k is the following function continuous at x = 2

 Prove that \(f(x) = \begin{cases} 2x+1,& \quad \text{}\text{whenx<2;}\\ k, & \quad \text{}\text{whenx=2;}\\ 3x-1, & \quad \text{}\text{whenx>2;} \end{cases} \) 

Answer 1

Since, f(x) is continuous at x = 2

k = 5