What is the extreme f(x)=(e^x)/(x^3) ?

The extreme f(x)=(e^x)/(x^3) is Minimum(3,(e^3)/(27))

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extreme f(x)=(e^x)/(x^3)

extreme points

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

Minimum (3, \frac{e ^{3}}{27})

Solution steps

f^{'} (x) = \frac{e ^{x} ( x - 3 )}{x ^{4}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 3

into

\frac{e ^{x}}{x ^{3}} : \frac{e ^{3}}{27}

Minimum (3, \frac{e ^{3}}{27})

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

q (x, y) = x^{2} - x y + 5 y^{2}

f (x, y) = 6 - x^{2} - 4 x - y^{2}

f (x) = 5 x^{3} - 4

f (x) = - x - \frac{144}{x} [9.13]

What is the extreme f(x)=(e^x)/(x^3) ?
The extreme f(x)=(e^x)/(x^3) is Minimum(3,(e^3)/(27))