What is the extreme f(x)=5+5x-5x^2,0<= x<= 3 ?

The extreme f(x)=5+5x-5x^2,0<= x<= 3 is Minimum(0,5),Maximum(1/2 , 25/4),Minimum(3,-25)

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extreme f(x)=5+5x-5x^2,0<= x<= 3

extreme points

f (x) = 5 + 5 x - 5 x^{2}, 0 \leq x \leq 3

Minimum (0, 5), Maximum (\frac{1}{2}, \frac{25}{4}), Minimum (3, - 25)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 3

Domain of

5 + 5 x - 5 x^{2}, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Increasing

: 0 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < 3

Plug the extreme point

x = 0

into

5 + 5 x - 5 x^{2} \Rightarrow y = 5

Minimum (0, 5)

Plug the extreme point

x = \frac{1}{2}

into

5 + 5 x - 5 x^{2} \Rightarrow y = \frac{25}{4}

Maximum (\frac{1}{2}, \frac{25}{4})

Plug the extreme point

x = 3

into

5 + 5 x - 5 x^{2} \Rightarrow y = - 25

Minimum (3, - 25)

Minimum (0, 5), Maximum (\frac{1}{2}, \frac{25}{4}), Minimum (3, - 25)

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

y = 1250 sin (2 x)

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

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

f (x) = - 4 x^{3} + 12 x^{2} - 10

What is the extreme f(x)=5+5x-5x^2,0<= x<= 3 ?
The extreme f(x)=5+5x-5x^2,0<= x<= 3 is Minimum(0,5),Maximum(1/2 , 25/4),Minimum(3,-25)