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(1,1)

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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, 1)

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 = 1

Minimum (1, 1)

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

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

f (x, y) = cos^{2} (t) \cdot sin^{2} (t)

f (x) = 7 x - 4 x ln (x)

f (x) = 2 x o n [1.64]

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(1,1)