Question

The minimum value of the function $y=2x^3-21x^2+36x-20$ is-

• A. $-128$
• B. $-126$
• C. $-120$
• D. None of these

Answer 1

The correct option is A $-128$

$y^{\prime }$
$=6x^2-42x+36=0$.. For critical point
Or
$6(x^2-7x+6)=0$
$6(x-1)(x-6)=0$
$x=1$ and $x=6$.
Now
$f"\left(x\right)$
$=12x-42$
For minima.
$f"\left(x\right)>0$
or
$x>\frac{7}{2}$
Or
$x>3.5$
Hence we get the minima at $x=6$.
Now
$f\left(6\right)$
$=2\left(216\right)-21\left(36\right)+36\left(6\right)-20$
$=-128$.