Let f:RR be a function defined by: [f(x)=casest^3-3t: t x, & x2 x^2+2x-6, & 2

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by:
$$f(x)=\begin{cases}\max\{t^3-3t: t\le x\}, & x\le2\ x^2+2x-6, & 2<x<3\ [x-3]+9, & 3\le x\le5\ 2x+1, & x>5\end{cases}$$
where $[t]$ is the greatest integer $\le t$. Let $m$ be the number of points where $f$ is not differentiable and $I=\int_{-2}^{2}f(x)\,dx$. Then the ordered pair $(m,I)$ is equal to:

Accepted Answer

For $x\le2$, $\max\{t^3-3t:t\le x\}$ works out to: $x^3-3x$ for $x\le-1$ (since $t^3-3t$ is increasing there) and $2$ for $-1<x\le2$ (the max value $2$ attained at $t=-1$ persists). So
$$f(x)=\begin{cases}x^3-3x, & x\le-1\2, & -1<x<2\x^2+2x-6, & 2<x<3\9, & 3\le x<4\10, & 4\le x<5\11, & x=5\2x+1, & x>5\end{cases}$$
Checking derivatives at the joins, $f$ is not differentiable at $x=2,3,4,5$, so $m=4$.
$$I=\int_{-2}^{-1}(x^3-3x)\,dx+\int_{-1}^{2}2\,dx=\frac{27}{4}$$
So $(m,I)=\left(4,\dfrac{27}{4}\right)$.

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