What is the extreme f(x)=x^4+4x^3-9,-4<= x<= 0 ?

The extreme f(x)=x^4+4x^3-9,-4<= x<= 0 is Maximum(-4,-9),Minimum(-3,-36),Maximum(0,-9)

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extreme f(x)=x^4+4x^3-9,-4<= x<= 0

extreme points

f (x) = x^{4} + 4 x^{3} - 9, - 4 \leq x \leq 0

Maximum (- 4, - 9), Minimum (- 3, - 36), Maximum (0, - 9)

Solution steps

Find the critical points:

x = - 4, x = - 3, x = 0

Domain of

x^{4} + 4 x^{3} - 9, - 4 \leq x \leq 0 : - 4 \leq x \leq 0

Find intervals:

Decreasing

: - 4 < x < - 3,

Increasing

: - 3 < x < 0

Plug the extreme point

x = - 4

into

x^{4} + 4 x^{3} - 9 \Rightarrow y = - 9

Maximum (- 4, - 9)

Plug the extreme point

x = - 3

into

x^{4} + 4 x^{3} - 9 \Rightarrow y = - 36

Minimum (- 3, - 36)

Plug the extreme point

x = 0

into

x^{4} + 4 x^{3} - 9 \Rightarrow y = - 9

Maximum (0, - 9)

Maximum (- 4, - 9), Minimum (- 3, - 36), Maximum (0, - 9)

f (x) = 3 x^{3} + \frac{3}{2} x^{2} - 2 x, - 1 \leq x \leq 1

2 - \frac{10 x}{x ^{2} + 25}

f (x) = 7 + 5 x - 5 x^{2}

f (x) = e^{- x} (x^{2} - 1)

h (x) = 2 (x - 5)^{3}

What is the extreme f(x)=x^4+4x^3-9,-4<= x<= 0 ?
The extreme f(x)=x^4+4x^3-9,-4<= x<= 0 is Maximum(-4,-9),Minimum(-3,-36),Maximum(0,-9)