Question

The absolute maximum value of a function $f$ given by
$f\left(x\right)=2x^3-15x^2+36x+1$ on the interval $[1,5]$ is

Answer 1

Given, $f\left(x\right)=2x^3-15x^2+36x+1$
$f^{\prime }\left(x\right)=6x^2-30x+36=6(x-3)(x-2)$
$\Rightarrow f^{\prime }\left(x\right)=0$
$\Rightarrow x=2,3$
Now, evaluating the value of $f$ at these points and the end points of the interval $[1,5],i.e.$ at $x=1,x=2,x=3,x=5$.
So,
$f\left(1\right)=24$
$f\left(2\right)=29$
$f\left(3\right)=28$
$f\left(5\right)=56$
Thus, absolute maximum value of $f$ on $[1,5]$ is $56$ occuring at $x=5.$