What is the extreme f(x)=(e^x)/(x^3),x>0 ?

The extreme f(x)=(e^x)/(x^3),x>0 is Minimum(3,(e^3)/(27))

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extreme f(x)=(e^x)/(x^3),x>0

extreme points

f (x) = \frac{e ^{x}}{x ^{3}}, x > 0

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

Solution steps

Find the critical points:

x = 3

Domain of

\frac{e ^{x}}{x ^{3}}, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug the extreme point

x = 3

into

\frac{e ^{x}}{x ^{3}} \Rightarrow y = \frac{e ^{3}}{27}

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

f (x, y) = ln (y - ∣ x ∣ - 1)

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

f (x) = 7 (x^{2} + 1)^{2} (x - 5)^{3}

f (x) = 9 x^{4} - 4 x^{3}, - 3 \leq x \leq 3

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

What is the extreme f(x)=(e^x)/(x^3),x>0 ?
The extreme f(x)=(e^x)/(x^3),x>0 is Minimum(3,(e^3)/(27))