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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = \frac{3}{2}

into

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

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

Plug the extreme point

x = 3

into

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

Minimum (3, 0)

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

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