What is the extreme f(x)=-x^3+12x-18 ?

The extreme f(x)=-x^3+12x-18 is Minimum(-2,-34),Maximum(2,-2)

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extreme f(x)=-x^3+12x-18

extreme points

f (x) = - x^{3} + 12 x - 18

Minimum (- 2, - 34), Maximum (2, - 2)

Solution steps

f^{'} (x) = - 3 x^{2} + 12

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

- x^{3} + 12 x - 18 : - 34

Minimum (- 2, - 34)

Plug

x = 2

into

- x^{3} + 12 x - 18 : - 2

Maximum (2, - 2)

Minimum (- 2, - 34), Maximum (2, - 2)

y = (x - 1)^{5}

f (x) = - x^{3} + 12 x - 25

f (x) = - x^{3} + 12 x - 23

f (x) = \frac{1}{3} x^{3} + 3 x^{2} + 5 x - \frac{11}{3}

f (x) = ∣ x ∣ + x^{2} - 5 x + 6

What is the extreme f(x)=-x^3+12x-18 ?
The extreme f(x)=-x^3+12x-18 is Minimum(-2,-34),Maximum(2,-2)