What is the extreme f(x)=6+3x-3x^2,0<= x<= 3 ?

The extreme f(x)=6+3x-3x^2,0<= x<= 3 is Minimum(0,6),Maximum(1/2 , 27/4),Minimum(3,-12)

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extreme f(x)=6+3x-3x^2,0<= x<= 3

extreme points

f (x) = 6 + 3 x - 3 x^{2}, 0 \leq x \leq 3

Minimum (0, 6), Maximum (\frac{1}{2}, \frac{27}{4}), Minimum (3, - 12)

Solution steps

Find the critical points:

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

Domain of

6 + 3 x - 3 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

6 + 3 x - 3 x^{2} \Rightarrow y = 6

Minimum (0, 6)

Plug the extreme point

x = \frac{1}{2}

into

6 + 3 x - 3 x^{2} \Rightarrow y = \frac{27}{4}

Maximum (\frac{1}{2}, \frac{27}{4})

Plug the extreme point

x = 3

into

6 + 3 x - 3 x^{2} \Rightarrow y = - 12

Minimum (3, - 12)

Minimum (0, 6), Maximum (\frac{1}{2}, \frac{27}{4}), Minimum (3, - 12)

\frac{x ^{2} + x}{x - 1}

\frac{ln ( x )}{6 x}

f (x) = 9 sin^{2} (x) + 9 sin (x)

h (x) = - 6 x^{3} + 18 x^{2} + 3

F (x) = \frac{2 x}{x ^{2} + 16}, - 8 \leq x \leq 8

What is the extreme f(x)=6+3x-3x^2,0<= x<= 3 ?
The extreme f(x)=6+3x-3x^2,0<= x<= 3 is Minimum(0,6),Maximum(1/2 , 27/4),Minimum(3,-12)