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

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

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extreme f(x)=(e^x)/(x^5)

extreme points

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

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

Solution steps

f^{'} (x) = \frac{e ^{x} ( x - 5 )}{x ^{6}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 5

into

\frac{e ^{x}}{x ^{5}} : \frac{e ^{5}}{3125}

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

y = 7 - 6 x - x^{3}

f (x) = \frac{1}{x ^{3}}

f (x) = x^{2} - 1, - 2 \leq x \leq 1

f (x) = - x^{3} + x^{2} + 8 x + 9

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

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