What is the extreme f(x)=-0.4x^2+90x-2000,0<= x ?

The extreme f(x)=-0.4x^2+90x-2000,0<= x is Minimum(0,-2000),Maximum(225/2 ,3062.5)

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extreme f(x)=-0.4x^2+90x-2000,0<= x

extreme points

f (x) = - 0.4 x^{2} + 90 x - 2000, 0 \leq x

Minimum (0, - 2000), Maximum (\frac{225}{2}, 3062.5)

Solution steps

Find the critical points:

x = 0, x = \frac{225}{2}

Domain of

- 0.4 x^{2} + 90 x - 2000, 0 \leq x : x \geq 0

Find intervals:

Increasing

: 0 < x < \frac{225}{2},

Decreasing

: \frac{225}{2} < x < \infty

Plug the extreme point

x = 0

into

- 0.4 x^{2} + 90 x - 2000 \Rightarrow y = - 2000

Minimum (0, - 2000)

Plug the extreme point

x = \frac{225}{2}

into

- 0.4 x^{2} + 90 x - 2000 \Rightarrow y = 3062.5

Maximum (\frac{225}{2}, 3062.5)

Minimum (0, - 2000), Maximum (\frac{225}{2}, 3062.5)

2 x^{2} - 2 x

f (x) = (x - 2)^{\frac{1}{8}}

y = 2 cos (x) - 11 x + 7 [- π, 0]

y = (x^{2} - 4)^{4} (x^{2} + 1)^{5}

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

What is the extreme f(x)=-0.4x^2+90x-2000,0<= x ?
The extreme f(x)=-0.4x^2+90x-2000,0<= x is Minimum(0,-2000),Maximum(225/2 ,3062.5)