Let f(x)=|2x^2+5|x|-3|,x R.

Question

Let $f(x)=|2{x}^{2}+5|x|-3|,x\in R$. If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m+n$ is equal to :

Accepted Answer

Let $u=|x|\geq0$; then $f(x)=2|u^2+5u-3|$. Since $u^2+5u-3$ is continuous in $u$ (hence in $x$) and the absolute value and doubling preserve continuity, $f$ is continuous everywhere: $m=0$.

Non-differentiability occurs wherever the inner expression $u^2+5u-3$ changes sign (a corner from the outer absolute value), i.e. at $u=u_0=\dfrac{-5+\sqrt{37}}{2}$ (the only non-negative root), giving $x=\pm u_0$ --- 2 points.

Additionally, near $x=0$: $u^2+5u-3=x^2+5|x|-3	o-3<0$ as $x	o0$, so locally $f(x)=2(3-5|x|-x^2)=6-10|x|-2x^2$, which has a corner at $x=0$ from the $-10|x|$ term --- a 3rd non-differentiable point.

So $n=3$, and $m+n=0+3=3$.

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