Question

If  \[f\left( x \right) = \begin{cases}2 x^2 + k, &\text{ if }  x \geq 0 \\ - 2 x^2 + k, & \text{ if }  x < 0\end{cases}\]  then what should be the value of k so that f(x) is continuous at x = 0.

 

Answer 1

The given function can be rewritten as 

\[f\left( x \right) = \begin{cases}2 x^2 + k, &\text{ if }  x \geq 0 \\ - 2 x^2 + k, & \text{ if }  x < 0\end{cases}\] 

We have
(LHL at x = 0) = 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} - 2 \left( - h \right)^2 + k = k\]

(RHL at x = 0) =\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( 2 h^2 + k \right) = k\]

If

\[f\left( x \right)\] is continuous at

\[x = 0\]

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = k\]

∴ k can be any real number.