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

The extreme f(x)=(4e^{4x})/(4x-5) is Minimum(3/2 ,4e^6)

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

extreme points

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

Minimum (\frac{3}{2}, 4 e^{6})

Solution steps

f^{'} (x) = \frac{4 ( 16 e ^{4 x} x - 24 e ^{4 x} )}{( 4 x - 5 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{5}{4},

Decreasing

: \frac{5}{4} < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = \frac{3}{2}

into

\frac{4 e ^{4 x}}{4 x - 5} : 4 e^{6}

Minimum (\frac{3}{2}, 4 e^{6})

f (x) = x + 3 + 6

f (x) = 2 x^{2} - 8 x + 5

f (x) = x^{4} (x - 1)^{2}

\frac{4 x ^{2} - 3 x ^{4}}{2 x}

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 54

What is the extreme f(x)=(4e^{4x})/(4x-5) ?
The extreme f(x)=(4e^{4x})/(4x-5) is Minimum(3/2 ,4e^6)