What is the extreme f(x)=9-4x-3x^2,-2<= x<= 1 ?

The extreme f(x)=9-4x-3x^2,-2<= x<= 1 is Minimum(-2,5),Maximum(-2/3 , 31/3),Minimum(1,2)

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extreme f(x)=9-4x-3x^2,-2<= x<= 1

extreme points

f (x) = 9 - 4 x - 3 x^{2}, - 2 \leq x \leq 1

Minimum (- 2, 5), Maximum (- \frac{2}{3}, \frac{31}{3}), Minimum (1, 2)

Solution steps

Find the critical points:

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

Domain of

9 - 4 x - 3 x^{2}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

x = - 2

into

9 - 4 x - 3 x^{2} \Rightarrow y = 5

Minimum (- 2, 5)

Plug the extreme point

x = - \frac{2}{3}

into

9 - 4 x - 3 x^{2} \Rightarrow y = \frac{31}{3}

Maximum (- \frac{2}{3}, \frac{31}{3})

Plug the extreme point

x = 1

into

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

Minimum (1, 2)

Minimum (- 2, 5), Maximum (- \frac{2}{3}, \frac{31}{3}), Minimum (1, 2)

f (x) = 3 x^{2} + 12 x - 9

f (x) = 395 x^{2} - 3160 x^{3}

e^{x - 1}

\frac{8 x ^{3}}{3} + 15 x^{2} - 8 x, - 5 \leq x \leq 1

f (x) = 2 x^{3} - 4 x - 1

What is the extreme f(x)=9-4x-3x^2,-2<= x<= 1 ?
The extreme f(x)=9-4x-3x^2,-2<= x<= 1 is Minimum(-2,5),Maximum(-2/3 , 31/3),Minimum(1,2)