Question

If \[f\left( x \right) = \begin{cases}\frac{x^2}{2}, & \text{ if } 0 \leq x \leq 1 \\ 2 x^2 - 3x + \frac{3}{2}, & \text P{ \text{ if }  }  1 < x \leq 2\end{cases}\]. Show that f is continuous at x = 1.

 

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{x^2}{2}, \text{ if } 0 \leq x \leq 1}{2 x^2 - 3x + \frac{3}{2}, if 1 < x \leq 2}\] 

We have
(LHL at x = 1) = 

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right)\]

\[= \lim_{h \to 0} \frac{\left( 1 - h \right)^2}{2} = \frac{1}{2}\]

(RHL at x = 1) =  

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right)\]

\[= \lim_{h \to 0} \left[ 2 \left( 1 + h \right)^2 - 3\left( 1 + h \right) + \frac{3}{2} \right] = 2 - 3 + \frac{3}{2} = \frac{1}{2}\]

Also,

\[f\left( 1 \right) = \frac{\left( 1 \right)^2}{2} = \frac{1}{2}\]

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]

Hence, the given function is continuous at

\[x = 1\]