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

The extreme f(x)=x^4-3x^2,-2<= x<= 0 is Maximum(-2,4),Minimum(-sqrt(3/2),-9/4),Maximum(0,0)

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extreme f(x)=x^4-3x^2,-2<= x<= 0

extreme points

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

Maximum (- 2, 4), Minimum (- \frac{3}{2}, - \frac{9}{4}), Maximum (0, 0)

Solution steps

Find the critical points:

x = - 2, x = - \frac{3}{2}, x = 0

Domain of

x^{4} - 3 x^{2}, - 2 \leq x \leq 0 : - 2 \leq x \leq 0

Find intervals:

Decreasing

: - 2 < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < 0

Plug the extreme point

x = - 2

into

x^{4} - 3 x^{2} \Rightarrow y = 4

Maximum (- 2, 4)

Plug the extreme point

x = - \frac{3}{2}

into

x^{4} - 3 x^{2} \Rightarrow y = - \frac{9}{4}

Minimum (- \frac{3}{2}, - \frac{9}{4})

Plug the extreme point

x = 0

into

x^{4} - 3 x^{2} \Rightarrow y = 0

Maximum (0, 0)

Maximum (- 2, 4), Minimum (- \frac{3}{2}, - \frac{9}{4}), Maximum (0, 0)

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

y = 15 x e^{- 0.05 x} - 30

f (x) = \frac{- 4 x}{x ^{2} + 6}, (0, 6)

f (x) = 2 x^{3} + 15 x^{2} - 36 x + 7, - 6 < x < 2

f (x) = 13 e^{- x}, (0, 4)

What is the extreme f(x)=x^4-3x^2,-2<= x<= 0 ?
The extreme f(x)=x^4-3x^2,-2<= x<= 0 is Maximum(-2,4),Minimum(-sqrt(3/2),-9/4),Maximum(0,0)