What is the extreme f(x)=0.05x+15+(320)/x ?

The extreme f(x)=0.05x+15+(320)/x is Maximum(-80,7),Minimum(80,23)

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extreme f(x)=0.05x+15+(320)/x

extreme points

f (x) = 0.05 x + 15 + \frac{320}{x}

Maximum (- 80, 7), Minimum (80, 23)

Solution steps

Find the critical points:

x = - 80, x = 80

Domain of

0.05 x + 15 + \frac{320}{x} : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 80,

Decreasing

: - 80 < x < 0,

Decreasing

: 0 < x < 80,

Increasing

: 80 < x < \infty

Plug the extreme point

x = - 80

into

0.05 x + 15 + \frac{320}{x} \Rightarrow y = 7

Maximum (- 80, 7)

Plug the extreme point

x = 80

into

0.05 x + 15 + \frac{320}{x} \Rightarrow y = 23

Minimum (80, 23)

Maximum (- 80, 7), Minimum (80, 23)

f (x) = x^{2} (5 - 4 x)^{2}

y = 4 x - 7, (0, - 6)

f (x) = ∣ x ∣

- lo g_{2} (x)

(- 3, \frac{1}{32}), (3, 128)

What is the extreme f(x)=0.05x+15+(320)/x ?
The extreme f(x)=0.05x+15+(320)/x is Maximum(-80,7),Minimum(80,23)