Find the absolute maximum and minimum values of a function $f$ given by

Solution:

Given : $f(x)=2 x^{3}-15 x^{2}+36 x+1$

$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$

For a local maximum or a local minimum, we have

$f^{\prime}(x)=0$

$\Rightarrow 6 x^{2}-30 x+36=0$

$\Rightarrow x^{2}-5 x+6=0$

$\Rightarrow(x-3)(x-2)=0$

$\Rightarrow x=2$ and $x=3$

Thus, the critical points of $f$ are $1,2,3$ and 5 .

Now,

$f(1)=2(1)^{3}-15(1)^{2}+36(1)+1=24$

$f(2)=2(2)^{3}-15(2)^{2}+36(2)+1=29$

$f(3)=2(3)^{3}-15(3)^{2}+36(3)+1=28$

$f(5)=2(5)^{3}-15(5)^{2}+36(5)+1=56$

Hence, the absolute maximum value when $x=5$ is 56 and the absolute minimum value when $x=1$ is 24 .