Prove that f(x) = {(5x - 4, when 0 < x ≤ 1),(4x^2 - 3x, when 1 < x < 2), is discontinuous at x = 0.

Question

Prove that \(f(x) = \begin{cases} 5x - 4, & \quad \text{when } 0 \text{ < x ≤ 1}\\ 4x^2 - 3x, & \quad \text{when } 1 \text{ < x < 2 } \end{cases} \) is continuous at x = 1

Answer 1

It is given that

\(f(x) = \begin{cases} 5x - 4, & \quad \text{when } 0 \text{ < x ≤ 1}\\ 4x^2 - 3x, & \quad \text{when } 1 \text{ < x < 2 } \end{cases} \)

Consider left hand limit at x = 1

Here the value of function at x = 1 is f(x) = 5x - 4 

We get f(1) = 5(1) - 4 = 1

So

Therefore, f(x) is continuous at x = 1.