What is the extreme f(x)=(x^3)/3-16x^2+240x-36 ?

The extreme f(x)=(x^3)/3-16x^2+240x-36 is Maximum(12,1116),Minimum(20, 3092/3)

extreme f(x)=(x^3)/3-16x^2+240x-36

extreme points

f (x) = \frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36

Maximum (12, 1116), Minimum (20, \frac{3092}{3})

Solution steps

f^{'} (x) = x^{2} - 32 x + 240

Find intervals:

Increasing

: - \infty < x < 12,

Decreasing

: 12 < x < 20,

Increasing

: 20 < x < \infty

Plug

x = 12

into

\frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36 : 1116

Maximum (12, 1116)

Plug

x = 20

into

\frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36 : \frac{3092}{3}

Minimum (20, \frac{3092}{3})

Maximum (12, 1116), Minimum (20, \frac{3092}{3})

y = 4 x^{3} - 8 x

f (x) = \frac{7}{4 - 3 sin ^{2} ( x )}

x^{2} - 5, 1 < x < 7

(x - 7) (x - 5)^{2} (x - 3) (x - 1)

\frac{400}{1 + 8 e ^{0.05 x}}

What is the extreme f(x)=(x^3)/3-16x^2+240x-36 ?
The extreme f(x)=(x^3)/3-16x^2+240x-36 is Maximum(12,1116),Minimum(20, 3092/3)