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

The extreme f(x)=2x^2+x+2,-1<= x<= 3 is Maximum(-1,3),Minimum(-1/4 , 15/8),Maximum(3,23)

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

extreme points

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

Maximum (- 1, 3), Minimum (- \frac{1}{4}, \frac{15}{8}), Maximum (3, 23)

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{4}, x = 3

Domain of

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

Find intervals:

Decreasing

: - 1 < x < - \frac{1}{4},

Increasing

: - \frac{1}{4} < x < 3

Plug the extreme point

x = - 1

into

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

Maximum (- 1, 3)

Plug the extreme point

x = - \frac{1}{4}

into

2 x^{2} + x + 2 \Rightarrow y = \frac{15}{8}

Minimum (- \frac{1}{4}, \frac{15}{8})

Plug the extreme point

x = 3

into

2 x^{2} + x + 2 \Rightarrow y = 23

Maximum (3, 23)

Maximum (- 1, 3), Minimum (- \frac{1}{4}, \frac{15}{8}), Maximum (3, 23)

f (x) = sin (x), 0 \leq x < π^{2}

\frac{- x ^{2} - x + 6}{x ^{2} + 3 x - 4}

f (x) = 2 x - 1 + 3

11 x

y = 2 e^{x} - 1

What is the extreme f(x)=2x^2+x+2,-1<= x<= 3 ?
The extreme f(x)=2x^2+x+2,-1<= x<= 3 is Maximum(-1,3),Minimum(-1/4 , 15/8),Maximum(3,23)