What is the extreme f(x)=(x^3)/3-5x^2+16x-100 ?

The extreme f(x)=(x^3)/3-5x^2+16x-100 is Maximum(2,-256/3),Minimum(8,-364/3)

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extreme f(x)=(x^3)/3-5x^2+16x-100

extreme points

f (x) = \frac{x ^{3}}{3} - 5 x^{2} + 16 x - 100

Maximum (2, - \frac{256}{3}), Minimum (8, - \frac{364}{3})

Solution steps

f^{'} (x) = x^{2} - 10 x + 16

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 2

into

\frac{x ^{3}}{3} - 5 x^{2} + 16 x - 100 : - \frac{256}{3}

Maximum (2, - \frac{256}{3})

Plug

x = 8

into

\frac{x ^{3}}{3} - 5 x^{2} + 16 x - 100 : - \frac{364}{3}

Minimum (8, - \frac{364}{3})

Maximum (2, - \frac{256}{3}), Minimum (8, - \frac{364}{3})

What is the extreme f(x)=(x^3)/3-5x^2+16x-100 ?
The extreme f(x)=(x^3)/3-5x^2+16x-100 is Maximum(2,-256/3),Minimum(8,-364/3)