Question
Let $M$ and $m$ be respectively the local maximum and the local minimum values of the function, $f(x) = \,2{x^3} - 9{x^2} + 12x + 5$ in the interval $[0, 3].$ Then $M-m$ is equal to
$\Rightarrow f^{\prime}(x)=6 x^{2}-18 x+12=0$
For maxima or minima put $f^{\prime}(x)=0$
$\Rightarrow x^{2}-3 x+2=0$
$\Rightarrow x=1$ or $x=2$
Now, $f^{\prime \prime}(x)=12 x-18$
$\Rightarrow f^{\prime \prime}(1)=12(1)-18=-6<0$
Hence, $f(x)$ has maxima at $x=1$
$\therefore$ maximum value
$=M=f(1)=2-9+12+5$ $=10$.
And, $f^{\prime \prime}(2)=12(2)-18=6>0$.
Hence, $f(x)$ has minima at $x=2$
$\therefore$ minimum value $=m=f(2)=2(8)-9(4)+$
$12 (2)$
$+5=9$
$\therefore M-m=10-9=1$
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$f(x)=\left\{\begin{array}{clr}\left|2 x^{2}-3 x-7\right| \, \text { if } x \leq-1 \\ {\left[4 x^{2}-1\right]} \text { if } -1 < x < 1 \\ |x+1|+|x-2| \text { if } x \geq 1\end{array}\right.$
$[t]$ denotes the greatest integer $\leq t$, is discontinuous is