What is the extreme f(x)=(5e^{4x})/(5x-3) ?

The extreme f(x)=(5e^{4x})/(5x-3) is Minimum(17/20 ,4e^{17/5})

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

extreme points

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

Minimum (\frac{17}{20}, 4 e^{\frac{17}{5}})

Solution steps

f^{'} (x) = \frac{5 ( 20 e ^{4 x} x - 17 e ^{4 x} )}{( 5 x - 3 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{3}{5},

Decreasing

: \frac{3}{5} < x < \frac{17}{20},

Increasing

: \frac{17}{20} < x < \infty

Plug

x = \frac{17}{20}

into

\frac{5 e ^{4 x}}{5 x - 3} : 4 e^{\frac{17}{5}}

Minimum (\frac{17}{20}, 4 e^{\frac{17}{5}})

f (x, y) = x y - x^{3} y^{2}

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

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

81 - x^{2}, - 9 \leq x \leq 9

\frac{( - 4 )}{x - 7}

What is the extreme f(x)=(5e^{4x})/(5x-3) ?
The extreme f(x)=(5e^{4x})/(5x-3) is Minimum(17/20 ,4e^{17/5})