What is the extreme f(x)=1296x-0.12x^3 ?

The extreme f(x)=1296x-0.12x^3 is Minimum(-60,-51840),Maximum(60,51840)

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extreme f(x)=1296x-0.12x^3

extreme points

f (x) = 1296 x - 0.12 x^{3}

Minimum (- 60, - 51840), Maximum (60, 51840)

Solution steps

f^{'} (x) = 1296 - 0.36 x^{2}

Find intervals:

Decreasing

: - \infty < x < - 60,

Increasing

: - 60 < x < 60,

Decreasing

: 60 < x < \infty

Plug

x = - 60

into

1296 x - 0.12 x^{3} : - 51840

Minimum (- 60, - 51840)

Plug

x = 60

into

1296 x - 0.12 x^{3} : 51840

Maximum (60, 51840)

Minimum (- 60, - 51840), Maximum (60, 51840)

f (x) = ln (x^{2} + 3 x + 7)

f (x) = x^{4} - x^{2} + 1

f (x) = x^{3} + 6 x^{2} - 15 x + 2

x^{3} + y^{3} - 6 x y

f (x) = \frac{2 x ^{2}}{x + 1}

What is the extreme f(x)=1296x-0.12x^3 ?
The extreme f(x)=1296x-0.12x^3 is Minimum(-60,-51840),Maximum(60,51840)