What is the extreme f(x)=x^3-6sqrt(x)-6 ?

The extreme f(x)=x^3-6sqrt(x)-6 is Maximum(0,-6),Minimum(1,-11)

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extreme f(x)=x^3-6sqrt(x)-6

extreme points

f (x) = x^{3} - 6 x - 6

Maximum (0, - 6), Minimum (1, - 11)

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

x^{3} - 6 x - 6 : x \geq 0

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = 0

into

x^{3} - 6 x - 6 \Rightarrow y = - 6

Maximum (0, - 6)

Plug the extreme point

x = 1

into

x^{3} - 6 x - 6 \Rightarrow y = - 11

Minimum (1, - 11)

Maximum (0, - 6), Minimum (1, - 11)

t^{4} - 4 t^{3} + 15

y = x (20 - 2 x) (16 - x)

y = e^{x} + 1

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

f (x) = 12 x^{3} - 24 x^{2} + 12 x

What is the extreme f(x)=x^3-6sqrt(x)-6 ?
The extreme f(x)=x^3-6sqrt(x)-6 is Maximum(0,-6),Minimum(1,-11)