What is the extreme f(x)=(2e^x)/(x^5) ?

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

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extreme f(x)=(2e^x)/(x^5)

extreme points

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

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

Solution steps

f^{'} (x) = \frac{2 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{2 e ^{x}}{x ^{5}} : \frac{2 e ^{5}}{3125}

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

f (x, y) = (9 x^{2} + 7 x + 10) (9 y^{2} + 4 y + 9)

f (x) = 4 x^{2} - 2 x + 1

a + (ln (x))^{2}

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

f (x, y) = x^{3} - 4 x^{2} - 4 y^{2} + 8 x y - 3 x

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