Determine the values for a for which $g$ is continuous & differentiable at all $x$ whereas $g(x)=(a+|x|)^2 e^{(5-|x|)^2}.$

Question

$g(x)=(a+|x|)^2 e^{(5-|x|)^2}.$

$g(x)=\begin{cases}(a-x)^2 e^{(5+x)^2} & x < 0\\(a+x)^2 e^{(5-x)^2} & x \geq 0\end{cases}$

Given $g $ is continuous at all $x$. Let $x=k$

$\lim_{x \rightarrow -k}g(x)=\lim_{x \rightarrow k}g(x)$

$\Rightarrow(a-k)^2 e^{(5+k)^2}=(a+k)^2 e^{(5-k)^2}$

& by solving this I got $a=k(\frac{e^{10k}+1}{e^{10k}-1}), k\neq0$

Likewise I solved for the derivative & got the equation

$(a-k)^2(5+k)e^{(5+k)^2}-(a-k)e^{(5+k)^2}=e^{(5-k)^2}(a+k)-e^{(5-k)^2}(a+k)^2(5-k)$

But I couldn't able to solve for $a$ here.

Answer 1

If $x \ne 0$, it is differentiable.

The point of interest is when $x=0$. $$g(0)=a^2\exp(5)$$

$$\lim_{x \to 0^+}g(x)=g(0)=\lim_{x\to 0^-}g(x)$$

Now, we investigate the differentiability,

\begin{align}\lim_{x \to 0^+}\frac{g(x)-g(0)}{x}&=\lim_{x \to 0^+}\frac{(a+x)^2e^{5-x^2}-a^2\exp(5)}{x}\\ &=\exp(5) \lim_{x \to 0^+}\frac{(a+x)^2\exp(-x^2)-a^2}{x}\\ &= \exp(5) \lim_{x \to 0^+}[2(a+x)\exp(-x^2)-2x(a+x)^2\exp(-x^2)]\\ &=2\exp(5)a\end{align}

Similarly,

\begin{align}\lim_{x \to 0^-}\frac{g(x)-g(0)}{x}&=\lim_{x \to 0^-}\frac{(a-x)^2e^{5+x^2}-a^2\exp(5)}{x}\\ &=\exp(5) \lim_{x \to 0^-}\frac{(a-x)^2\exp(x^2)-a^2}{x}\\ &= \exp(5) \lim_{x \to 0^-}[-2(a-x)\exp(x^2)+2x(a-x)^2\exp(x^2)]\\ &=-2\exp(5)a\end{align}

Hence, to make it differentiable, $a$ has to be $0$.