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

The extreme f(x)=2x^2+3,-1<= x<= 2 is Maximum(-1,5),Minimum(0,3),Maximum(2,11)

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

extreme points

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

Maximum (- 1, 5), Minimum (0, 3), Maximum (2, 11)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 2

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 1

into

2 x^{2} + 3 \Rightarrow y = 5

Maximum (- 1, 5)

Plug the extreme point

x = 0

into

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

Minimum (0, 3)

Plug the extreme point

x = 2

into

2 x^{2} + 3 \Rightarrow y = 11

Maximum (2, 11)

Maximum (- 1, 5), Minimum (0, 3), Maximum (2, 11)

f (x) = sin (π x) (- \frac{1}{3}, \frac{1}{3})

f (x) = - 2 x^{2} (x + 6)

f (x, y) = 3 x^{2} - 6 x + 4 y^{2} + 8

f (x) = 2 x^{3} + 11 x^{2} - 8 x - 12

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

What is the extreme f(x)=2x^2+3,-1<= x<= 2 ?
The extreme f(x)=2x^2+3,-1<= x<= 2 is Maximum(-1,5),Minimum(0,3),Maximum(2,11)