What is the extreme f(x)=x^4-6x^2+9,0<= x<= 2 ?

The extreme f(x)=x^4-6x^2+9,0<= x<= 2 is Maximum(0,9),Minimum(sqrt(3),0),Maximum(2,1)

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extreme f(x)=x^4-6x^2+9,0<= x<= 2

extreme points

f (x) = x^{4} - 6 x^{2} + 9, 0 \leq x \leq 2

Maximum (0, 9), Minimum (3, 0), Maximum (2, 1)

Solution steps

Find the critical points:

x = 0, x = 3, x = 2

Domain of

x^{4} - 6 x^{2} + 9, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < 2

Plug the extreme point

x = 0

into

x^{4} - 6 x^{2} + 9 \Rightarrow y = 9

Maximum (0, 9)

Plug the extreme point

x = 3

into

x^{4} - 6 x^{2} + 9 \Rightarrow y = 0

Minimum (3, 0)

Plug the extreme point

x = 2

into

x^{4} - 6 x^{2} + 9 \Rightarrow y = 1

Maximum (2, 1)

Maximum (0, 9), Minimum (3, 0), Maximum (2, 1)

\frac{4 - x ^{2}}{x}

f (x) = e^{- k x}

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

f (x, y) = x^{3} - y^{2} - 12 x + 6 y

f (x) = x^{4} - 8 x^{3} + 4

What is the extreme f(x)=x^4-6x^2+9,0<= x<= 2 ?
The extreme f(x)=x^4-6x^2+9,0<= x<= 2 is Maximum(0,9),Minimum(sqrt(3),0),Maximum(2,1)