If f(x) = {[x^2/2, if 0≤x≤1][2x^2-3x+3/2,if 1<x≤ 2Show that f is continuous at x = 1. ​

Question

If \(f(x) = \begin{cases}\frac{x^2}{2}&,\quad if\,0≤x≤1\\2x^2-3x+\frac{3}{2}&,\quad if\,1<x≤2\end{cases} \)

Show that f is continuous at x = 1.

Answer 1

Given : 

For f(x) is continuous at x = 1 

If f(x) to be continuous at x = 1,we have to show, 

f(1)^– = f(1)⁺ = f(1) 

LHL = f(1)^– = \(\lim\limits_{x \to 1}(\frac{x^2}{2})\)

⇒ \(\frac{1}{2}\) ...(1)

⇒ \(\frac{1}{2}\) ...(2)

From (1) & (2),we get 

f(1)^– = f(1)⁺ 

Hence, 

f(x) is continuous at x = 1