What is the extreme f(x)=(x^3)/3-10x^2+96x+100 ?

The extreme f(x)=(x^3)/3-10x^2+96x+100 is Maximum(8, 1196/3),Minimum(12,388)

extreme f(x)=(x^3)/3-10x^2+96x+100

extreme points

f (x) = \frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100

Maximum (8, \frac{1196}{3}), Minimum (12, 388)

Solution steps

f^{'} (x) = x^{2} - 20 x + 96

Find intervals:

Increasing

: - \infty < x < 8,

Decreasing

: 8 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = 8

into

\frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100 : \frac{1196}{3}

Maximum (8, \frac{1196}{3})

Plug

x = 12

into

\frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100 : 388

Minimum (12, 388)

Maximum (8, \frac{1196}{3}), Minimum (12, 388)

f (x) = - x^{3} + 9 x^{2} + 120 x - 200, x \geq 5

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

f (x) = x^{\frac{2}{3}} (x - 6)

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

f (x) = - 2 x^{2} - 5 x y - 6 y^{2} + 32 x + 63 y + 9

What is the extreme f(x)=(x^3)/3-10x^2+96x+100 ?
The extreme f(x)=(x^3)/3-10x^2+96x+100 is Maximum(8, 1196/3),Minimum(12,388)