What is the extreme f(x)=2x^3+45x^2-300x,4<= x<= 11 ?

The extreme f(x)=2x^3+45x^2-300x,4<= x<= 11 is Minimum(4,-352),Maximum(11,4807)

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extreme f(x)=2x^3+45x^2-300x,4<= x<= 11

extreme points

f (x) = 2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11

Minimum (4, - 352), Maximum (11, 4807)

Solution steps

Find the critical points:

x = 4, x = 11

Domain of

2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11 : 4 \leq x \leq 11

Find intervals:

Increasing

: 4 < x < 11

Plug the extreme point

x = 4

into

2 x^{3} + 45 x^{2} - 300 x \Rightarrow y = - 352

Minimum (4, - 352)

Plug the extreme point

x = 11

into

2 x^{3} + 45 x^{2} - 300 x \Rightarrow y = 4807

Maximum (11, 4807)

Minimum (4, - 352), Maximum (11, 4807)

f (x) = 2 x^{4} - x^{2}

f (x) = x^{2} + \frac{1}{x}

p (x) = 12 x^{4} + 15 x^{3} + a x^{2} - 69 x - 30

f (x, y) = x + 2 \frac{y}{2} x + y

f (x, y) = 16 - (x - 2)^{2} - (y - 2)^{2}

What is the extreme f(x)=2x^3+45x^2-300x,4<= x<= 11 ?
The extreme f(x)=2x^3+45x^2-300x,4<= x<= 11 is Minimum(4,-352),Maximum(11,4807)