Question

Let $f\left(x\right)=\left|\right(x-1\left)(x^2-2x-3)\right|+x-3,x\in R$. If $m\text{and}M$ are respectively the number of points of local minimum and local maximum of $f$ in the interval $(0,4)$, then $m+M$ is equal to

Answer 1

$f\left(x\right)=|(x-1)(x+1)(x-3)|+(x-3)$

$f\left(x\right)=\{\begin{array}{cc}(x-3)\left(x^2\right)& 3\le x<4\\ (x-3)(2-x^2)& 1\le x<3\\ (x-3)\left(x^2\right)& 0<x<1\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2-6x& 3<x<4\\ -3x^2+6x+2& 1<x<3\\ 3x^2-6x& 0<x<1\end{array}$

$f^{\prime }\left(3^{+}\right)>0,f^{\prime }\left(3^{-}\right)<0\to$ Minimum

$f^{\prime }\left(1^{+}\right)>0,f^{\prime }\left(1^{-}\right)<0\to$ Minimum

for $x\in (1,3),f^{\prime }\left(x\right)=0$ at one point, $x=1+\sqrt{\frac{5}{3}}$ and $f^{\prime \prime }\left(x\right)<0$ $\to$ Maximum
for $x\in (3,4),f^{\prime }\left(x\right)\ne 0$
for $x\in (0,1),f^{\prime }\left(x\right)\ne 0$
So, $m+M=2+1=3$