What is the extreme f(x)=400x^2-1600x^3,0<= x<= 0.25 ?

The extreme f(x)=400x^2-1600x^3,0<= x<= 0.25 is Minimum(0,0),Maximum(1/6 , 100/27),Minimum(0.25,0)

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extreme f(x)=400x^2-1600x^3,0<= x<= 0.25

extreme points

f (x) = 400 x^{2} - 1600 x^{3}, 0 \leq x \leq 0.25

Minimum (0, 0), Maximum (\frac{1}{6}, \frac{100}{27}), Minimum (0.25, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{6}, x = 0.25

Domain of

400 x^{2} - 1600 x^{3}, 0 \leq x \leq 0.25 : 0 \leq x \leq 0.25

Find intervals:

Increasing

: 0 < x < \frac{1}{6},

Decreasing

: \frac{1}{6} < x < 0.25

Plug the extreme point

x = 0

into

400 x^{2} - 1600 x^{3} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{6}

into

400 x^{2} - 1600 x^{3} \Rightarrow y = \frac{100}{27}

Maximum (\frac{1}{6}, \frac{100}{27})

Plug the extreme point

x = 0.25

into

400 x^{2} - 1600 x^{3} \Rightarrow y = 0

Minimum (0.25, 0)

Minimum (0, 0), Maximum (\frac{1}{6}, \frac{100}{27}), Minimum (0.25, 0)

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

y = \frac{1}{2} x^{4} - 4 x^{2} + 3

y = - 4 x^{3} - 6 x^{2} + 360 x - 12

f (x) = 2 x^{4} + x^{3} - 13 x^{2} - 9 x - 45

What is the extreme f(x)=400x^2-1600x^3,0<= x<= 0.25 ?
The extreme f(x)=400x^2-1600x^3,0<= x<= 0.25 is Minimum(0,0),Maximum(1/6 , 100/27),Minimum(0.25,0)