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

The extreme f(x)=(e^{3x})/(3x-3) is Minimum(4/3 ,e^4)

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extreme f(x)=(e^{3x})/(3x-3)

extreme points

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

Minimum (\frac{4}{3}, e^{4})

Solution steps

f^{'} (x) = \frac{e ^{3 x} ( 3 x - 4 )}{3 ( x - 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 1,

Decreasing

: 1 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = \frac{4}{3}

into

\frac{e ^{3 x}}{3 x - 3} : e^{4}

Minimum (\frac{4}{3}, e^{4})

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

f (x) = \frac{1}{4} x^{4} + \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4

f (x) = x y^{2} = 9

f (x) = sin (θ)

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

What is the extreme f(x)=(e^{3x})/(3x-3) ?
The extreme f(x)=(e^{3x})/(3x-3) is Minimum(4/3 ,e^4)