What is the extreme f(x)=(e^x)/(x^5),x>0 ?

The extreme f(x)=(e^x)/(x^5),x>0 is Minimum(5,(e^5)/(3125))

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extreme f(x)=(e^x)/(x^5),x>0

extreme points

f (x) = \frac{e ^{x}}{x ^{5}}, x > 0

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

Solution steps

Find the critical points:

x = 5

Domain of

\frac{e ^{x}}{x ^{5}}, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug the extreme point

x = 5

into

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

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

f (x) = 8 + 3 x^{2} - x^{3}

f (x) = 4 x^{2} - 16 x + 4, 0 \leq x \leq 3

f (x, y) = 2 x^{2} + 4 x + 2 y^{2} + 5 y

f (x) = x^{3} - 2^{2}

- x^{3} + 5 x^{2} + 8 x + 9

What is the extreme f(x)=(e^x)/(x^5),x>0 ?
The extreme f(x)=(e^x)/(x^5),x>0 is Minimum(5,(e^5)/(3125))