What is the extreme f(x)=(2e^{2x})/(5x-15) ?

The extreme f(x)=(2e^{2x})/(5x-15) is Minimum(7/2 ,(4e^7)/5)

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extreme f(x)=(2e^{2x})/(5x-15)

extreme points

f (x) = \frac{2 e ^{2 x}}{5 x - 15}

Minimum (\frac{7}{2}, \frac{4 e ^{7}}{5})

Solution steps

f^{'} (x) = \frac{2 e ^{2 x} ( 2 x - 7 )}{5 ( x - 3 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 3,

Decreasing

: 3 < x < \frac{7}{2},

Increasing

: \frac{7}{2} < x < \infty

Plug

x = \frac{7}{2}

into

\frac{2 e ^{2 x}}{5 x - 15} : \frac{4 e ^{7}}{5}

Minimum (\frac{7}{2}, \frac{4 e ^{7}}{5})

f (x) = x^{3} + 12 x^{2} - 27 x + 2, - 10 \leq x \leq 2

f (x) = 200 x - \frac{2}{3} x^{2}

2 x^{2} y + x y^{2} - x

f (x) = x + x - 1

f (x) = x + 4 y

What is the extreme f(x)=(2e^{2x})/(5x-15) ?
The extreme f(x)=(2e^{2x})/(5x-15) is Minimum(7/2 ,(4e^7)/5)