What is the extreme f(x)=-x^2+3x,-1<= x<= 3 ?

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

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extreme f(x)=-x^2+3x,-1<= x<= 3

extreme points

f (x) = - x^{2} + 3 x, - 1 \leq x \leq 3

Minimum (- 1, - 4), Maximum (\frac{3}{2}, \frac{9}{4}), Minimum (3, 0)

Solution steps

Find the critical points:

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

Domain of

- x^{2} + 3 x, - 1 \leq x \leq 3 : - 1 \leq x \leq 3

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

x = - 1

into

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

Minimum (- 1, - 4)

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

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

\frac{x ^{2} - x - 6}{x - 2}

f (x) = \frac{x - 1}{x}

\frac{1}{6 x - 12} 2 x + 16

f (x) = 4 x^{2} - 2 x, (1, 4)

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