Question

Examine the maxima and minima of the function $f\left(x\right)=2x^3-21x^2+36x-20$. Also, find the maximum and minimum values of $f\left(x\right)$.

Answer 1

Given, $f\left(x\right)=2x^3-21x^2+36x-20$
$\therefore f^{\prime }\left(x\right)=6x^2-42x+36$
When $f\left(x\right)$ is a maximum or a minimum, $f^{\prime }\left(x\right)=0$
Hence, $6x^2-42x+36=0$
$x^2-7x+6=0$
$x^2-6x-x+6=0$
$x(x-6)-1(x-6)=0$
$(x-6)(x-1)=0$
$x=1,6$
Again $f^{\prime \prime }\left(x\right)=12x-42$
$=6(2x-7)$
Now, when $x=1,f^{\prime \prime }\left(x\right)=-30$ ....[negative]
And when $x=6,f^{\prime \prime }\left(x\right)=30$ ....[positive]
Hence, $f\left(x\right)$ is maximum for $x=1$ and minimum for $x=6$.
The maximum and minimum values of $f\left(x\right)$ are
$f\left(1\right)=2(1{)}^3-21(1{)}^2+36\left(1\right)-20$
$=2-21+36-20=-3$
$f\left(6\right)=2(6{)}^2-21(6{)}^2+36\left(6\right)-20$
$=432-756+216-20=-128$