What is the extreme f(x)=(e^{2x})/(x-3) ?

The extreme f(x)=(e^{2x})/(x-3) is Minimum(7/2 ,2e^7)

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extreme f(x)=(e^{2x})/(x-3)

extreme points

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

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

Solution steps

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

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

f (x) = 8 - x, - 1 \leq x \leq 3

P (x, y) = x^{2} + 5 x y - 4 y^{2}

f (x) = 2 x^{3} - 6 x^{2} - 18 x + 2

f (x) = cos (3 x), - \frac{1}{6} π < x < \frac{1}{2} π

f (x) = 8 x^{3}

What is the extreme f(x)=(e^{2x})/(x-3) ?
The extreme f(x)=(e^{2x})/(x-3) is Minimum(7/2 ,2e^7)