What is the extreme f(x)=5+54x+2x^3,0<= x<= 4 ?

The extreme f(x)=5+54x+2x^3,0<= x<= 4 is Minimum(0,5),Maximum(4,349)

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extreme f(x)=5+54x+2x^3,0<= x<= 4

extreme points

f (x) = 5 + 54 x + 2 x^{3}, 0 \leq x \leq 4

Minimum (0, 5), Maximum (4, 349)

Solution steps

Find the critical points:

x = 0, x = 4

Domain of

5 + 54 x + 2 x^{3}, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Increasing

: 0 < x < 4

Plug the extreme point

x = 0

into

5 + 54 x + 2 x^{3} \Rightarrow y = 5

Minimum (0, 5)

Plug the extreme point

x = 4

into

5 + 54 x + 2 x^{3} \Rightarrow y = 349

Maximum (4, 349)

Minimum (0, 5), Maximum (4, 349)

f (x) = x^{3} - 4 x^{2} + 5 x

f (x) = \frac{x ^{3}}{3} - 4 x^{2} + 12 x, 0 \leq x \leq 7

f (x, y) = x^{2} + y^{2} - 1

f (x) = 2 x^{3} - 12 x^{2} + 18 x - 5

x^{2} + 4 x y + y^{2}

What is the extreme f(x)=5+54x+2x^3,0<= x<= 4 ?
The extreme f(x)=5+54x+2x^3,0<= x<= 4 is Minimum(0,5),Maximum(4,349)