What is the extreme f(x)=(x^2+3x+9),-2<= x<= 1 ?

The extreme f(x)=(x^2+3x+9),-2<= x<= 1 is Maximum(-2,7),Minimum(-3/2 , 27/4),Maximum(1,13)

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extreme f(x)=(x^2+3x+9),-2<= x<= 1

extreme points

f (x) = (x^{2} + 3 x + 9), - 2 \leq x \leq 1

Maximum (- 2, 7), Minimum (- \frac{3}{2}, \frac{27}{4}), Maximum (1, 13)

Solution steps

Find the critical points:

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

Domain of

x^{2} + 3 x + 9, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

x = - 2

into

x^{2} + 3 x + 9 \Rightarrow y = 7

Maximum (- 2, 7)

Plug the extreme point

x = - \frac{3}{2}

into

x^{2} + 3 x + 9 \Rightarrow y = \frac{27}{4}

Minimum (- \frac{3}{2}, \frac{27}{4})

Plug the extreme point

x = 1

into

x^{2} + 3 x + 9 \Rightarrow y = 13

Maximum (1, 13)

Maximum (- 2, 7), Minimum (- \frac{3}{2}, \frac{27}{4}), Maximum (1, 13)

2 x + \frac{2}{x}

U (x, y) = 2 x (1 - y)

f (x) = 4 x^{3} - x^{2} - x + 8

f (x) = - 3 sec (x)

x x - 2

What is the extreme f(x)=(x^2+3x+9),-2<= x<= 1 ?
The extreme f(x)=(x^2+3x+9),-2<= x<= 1 is Maximum(-2,7),Minimum(-3/2 , 27/4),Maximum(1,13)