Question

If $f\left(x\right)=\left\{\begin{array}{ll}2x^2+k,& \mathrm{if}x\ge 0\\ -2x^2+k,& \mathrm{if}x<0\end{array}\right.$, 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

$f\left(x\right)=\left\{\begin{array}{l}2x^2+k,\mathrm{if}x\ge 0\\ -2x^2+k,\mathrm{if}x<0\end{array}\right.$
​
We have
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }-2{\left(-h\right)}^2+k=k$

(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }f\left(h\right)=\underset{h\to 0}{\lim }\left(2h^2+k\right)=k$

If $f\left(x\right)$ is continuous at $x=0$, then

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=k \end{gathered}$$

∴ k can be any real number.