Let f(x)= array|c|c| e^x-1 & ,x

Question

Let $f(x)= \begin{array}{|c|c|} {e}^{x-1} & ,x<0 \ {x}^{2}-5x+6 & ,x\ge 0 \end{array}$ and $g(x)=f(|x|)+|f(x)|$. If the number of points where $g$ is not continuous and is not differentiable are $\alpha $ and $\beta $ respectively, then $\alpha +\beta $ is equal to $____$

Answer: 4

Accepted Answer

First look at $|f(x)|$.

For $xFor $x\, \, ge0$, $f(x)={x}\^{2}-5x+6=(x-2)(x-3)$. This becomes $0$ at $x=2$ and $x=3$, and it changes sign between $2$ and $3$. So when we take absolute value, $|f(x)|$ gets a ``sharp corner'' at $x=2$ and $x=3$, hence it is not differentiable at $x=2,3$.

At $x=0$, check continuity of $|f(x)|$: left value tends to $|{e}\^{-1}|={e}\^{-1}$, but right value is $|f(0)|=|6|=6$. Since these are not equal, $|f(x)|$ is discontinuous at $x=0$ (and therefore also not differentiable at $x=0$).

Now consider $g(x)=f(|x|)+|f(x)|$.

Here $f(|x|)$ always uses the $x\, \, ge0$ definition because $|x|\, \, ge0$. So $f(|x|)$ is continuous for all $x$, but it is not differentiable at $x=0$ because of $|x|$.

So, discontinuity in $g$ can only come from $|f(x)|$, which happens at $x=0$ only. Also, $g$ will be non-differentiable where either term is non-differentiable, i.e., at $x=0,2,3$.

$$$\begin{gathered} |f(x)|x=0 \ |f(x)|x=0,2,3 \ g(x)=f(|x|)+|f(x)| \ g(x)x=0 \ g(x)x=0,2,3 \ \alpha =1,\beta =3 \ \alpha +\beta =1+3=4 \end{gathered}$$

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