What is the extreme f(x)=2x^2+3x-5,-4<= x<= 4 ?

The extreme f(x)=2x^2+3x-5,-4<= x<= 4 is Maximum(-4,15),Minimum(-3/4 ,-49/8),Maximum(4,39)

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extreme f(x)=2x^2+3x-5,-4<= x<= 4

extreme

Maximum (- 4, 15), Minimum (- \frac{3}{4}, - \frac{49}{8}), Maximum (4, 39)

Solution steps

Find the critical points:

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

Domain of

2 x^{2} + 3 x - 5, - 4 \leq x \leq 4 : - 4 \leq x \leq 4

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

x = - 4

into

2 x^{2} + 3 x - 5 \Rightarrow y = 15

Maximum (- 4, 15)

Plug the extreme point

x = - \frac{3}{4}

into

2 x^{2} + 3 x - 5 \Rightarrow y = - \frac{49}{8}

Minimum (- \frac{3}{4}, - \frac{49}{8})

Plug the extreme point

x = 4

into

2 x^{2} + 3 x - 5 \Rightarrow y = 39

Maximum (4, 39)

Maximum (- 4, 15), Minimum (- \frac{3}{4}, - \frac{49}{8}), Maximum (4, 39)

What is the extreme f(x)=2x^2+3x-5,-4<= x<= 4 ?
The extreme f(x)=2x^2+3x-5,-4<= x<= 4 is Maximum(-4,15),Minimum(-3/4 ,-49/8),Maximum(4,39)