The minimum value of the function

Solution:

(a) $-128$

Given : $f(x)=2 x^{3}-21 x^{2}+36 x-20$

$\Rightarrow f^{\prime}(x)=6 x^{2}-42 x+36$

For a local maxima or a local minima, we must have

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

$\Rightarrow 6 x^{2}-42 x+36=0$

$\Rightarrow x^{2}-7 x+6=0$

$\Rightarrow(x-1)(x-6)=0$

$\Rightarrow x=1,6$

Now,

$f^{\prime \prime}(x)=12 x-42$

$\Rightarrow f^{\prime \prime}(1)=12-42=-30<0$

So, $x=1$ is a local maxima.

Also,

$f^{\prime \prime}(6)=72-42=30>0$

So, $x=6$ is a local miniima.

The local minimum value is given by

$f(6)==2(6)^{3}-21(6)^{2}+36(6)-20=-128$