What is the extreme f(x)=8x^2+20x+8,1<= x<= 3 ?

The extreme f(x)=8x^2+20x+8,1<= x<= 3 is Minimum(1,36),Maximum(3,140)

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extreme f(x)=8x^2+20x+8,1<= x<= 3

extreme points

f (x) = 8 x^{2} + 20 x + 8, 1 \leq x \leq 3

Minimum (1, 36), Maximum (3, 140)

Solution steps

Find the critical points:

x = 1, x = 3

Domain of

8 x^{2} + 20 x + 8, 1 \leq x \leq 3 : 1 \leq x \leq 3

Find intervals:

Increasing

: 1 < x < 3

Plug the extreme point

x = 1

into

8 x^{2} + 20 x + 8 \Rightarrow y = 36

Minimum (1, 36)

Plug the extreme point

x = 3

into

8 x^{2} + 20 x + 8 \Rightarrow y = 140

Maximum (3, 140)

Minimum (1, 36), Maximum (3, 140)

f (x) = \frac{1}{6} x^{3} - 8 x + \frac{19}{3}

f (x) = x - ln (6 x)

f (x, y) = x y^{2} - x y - x^{2} y

- 7 x^{2} + 126 x - 560

20 - 0.06 x + 0.0002 x^{2}

What is the extreme f(x)=8x^2+20x+8,1<= x<= 3 ?
The extreme f(x)=8x^2+20x+8,1<= x<= 3 is Minimum(1,36),Maximum(3,140)