What is the extreme f(x)=4+81x-3x^3,0<= x<= 4 ?

The extreme f(x)=4+81x-3x^3,0<= x<= 4 is Minimum(0,4),Maximum(3,166),Minimum(4,136)

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extreme f(x)=4+81x-3x^3,0<= x<= 4

extreme points

f (x) = 4 + 81 x - 3 x^{3}, 0 \leq x \leq 4

Minimum (0, 4), Maximum (3, 166), Minimum (4, 136)

Solution steps

Find the critical points:

x = 0, x = 3, x = 4

Domain of

4 + 81 x - 3 x^{3}, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < 4

Plug the extreme point

x = 0

into

4 + 81 x - 3 x^{3} \Rightarrow y = 4

Minimum (0, 4)

Plug the extreme point

x = 3

into

4 + 81 x - 3 x^{3} \Rightarrow y = 166

Maximum (3, 166)

Plug the extreme point

x = 4

into

4 + 81 x - 3 x^{3} \Rightarrow y = 136

Minimum (4, 136)

Minimum (0, 4), Maximum (3, 166), Minimum (4, 136)

f (x) = x^{2} - 6 x + 13

y = 2 x^{3} - 12 x^{2} + 9

\frac{8}{x} + 2 π x^{2}

f (x, y) = 8 x \frac{5}{6} y \frac{1}{6}

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

What is the extreme f(x)=4+81x-3x^3,0<= x<= 4 ?
The extreme f(x)=4+81x-3x^3,0<= x<= 4 is Minimum(0,4),Maximum(3,166),Minimum(4,136)