What is the extreme f(x)=(2e^x)/(x^5),[0,infinity ] ?

The extreme f(x)=(2e^x)/(x^5),[0,infinity ] is Minimum(5,(2e^5)/(3125))

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extreme f(x)=(2e^x)/(x^5),[0,infinity ]

extreme points

f (x) = \frac{2 e ^{x}}{x ^{5}}, [0, \infty]

Minimum (5, \frac{2 e ^{5}}{3125})

Solution steps

Find the critical points:

x = 5

Domain of

\frac{2 e ^{x}}{x ^{5}}, [0, \infty] : x > 0

Find intervals:

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug the extreme point

x = 5

into

\frac{2 e ^{x}}{x ^{5}} \Rightarrow y = \frac{2 e ^{5}}{3125}

Minimum (5, \frac{2 e ^{5}}{3125})

5 x^{3} e^{- x}

f (x) = x - 2 sin (x), (0, 2 π)

f (x) = \frac{- 1}{5 x ^{2} + 25 x + 30}

f (x) = \frac{5 x}{x ^{2} - 1}

f (x) = - \frac{x ^{3}}{3} + 2 x^{2} + 5 x + 1

What is the extreme f(x)=(2e^x)/(x^5),[0,infinity ] ?
The extreme f(x)=(2e^x)/(x^5),[0,infinity ] is Minimum(5,(2e^5)/(3125))