What is the extreme f(x)=x^3-12x^2+1 ?

The extreme f(x)=x^3-12x^2+1 is Maximum(0,1),Minimum(8,-255)

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extreme f(x)=x^3-12x^2+1

extreme points

f (x) = x^{3} - 12 x^{2} + 1

Maximum (0, 1), Minimum (8, - 255)

Solution steps

f^{'} (x) = 3 x^{2} - 24 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

x^{3} - 12 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 8

into

x^{3} - 12 x^{2} + 1 : - 255

Minimum (8, - 255)

Maximum (0, 1), Minimum (8, - 255)

f (x) = x^{2} + 4 x + 6

1 - x^{\frac{2}{3}}

f (x, y) = x^{2} - 0.5 y^{2} + 3 x

f (x) = 8 sin (∣ x ∣), - 2 π \leq x \leq 2 π

18 cos (θ) + 9 sin^{2} (θ), - π \leq θ \leq π

What is the extreme f(x)=x^3-12x^2+1 ?
The extreme f(x)=x^3-12x^2+1 is Maximum(0,1),Minimum(8,-255)