What is the extreme f(x)=(x^3)/3-5x^2-4 ?

The extreme f(x)=(x^3)/3-5x^2-4 is Maximum(0,-4),Minimum(10,-512/3)

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extreme f(x)=(x^3)/3-5x^2-4

extreme points

f (x) = \frac{x ^{3}}{3} - 5 x^{2} - 4

Maximum (0, - 4), Minimum (10, - \frac{512}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 10,

Increasing

: 10 < x < \infty

Plug

x = 0

into

\frac{x ^{3}}{3} - 5 x^{2} - 4 : - 4

Maximum (0, - 4)

Plug

x = 10

into

\frac{x ^{3}}{3} - 5 x^{2} - 4 : - \frac{512}{3}

Minimum (10, - \frac{512}{3})

Maximum (0, - 4), Minimum (10, - \frac{512}{3})

f (x) = x^{3} - 5 x + 2

f (x) = (x - 3) e^{x} + 3

x^{4} + x^{2}

f (x) = x^{7} - 4 x^{5} + 10

f (x) = x^{2} + 16

What is the extreme f(x)=(x^3)/3-5x^2-4 ?
The extreme f(x)=(x^3)/3-5x^2-4 is Maximum(0,-4),Minimum(10,-512/3)