What is the extreme f(x)=(5e^{4x})/(2x-5) ?

The extreme f(x)=(5e^{4x})/(2x-5) is Minimum(11/4 ,10e^{11})

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extreme f(x)=(5e^{4x})/(2x-5)

extreme points

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

Minimum (\frac{11}{4}, 10 e^{11})

Solution steps

f^{'} (x) = \frac{5 ( 8 e ^{4 x} x - 22 e ^{4 x} )}{( 2 x - 5 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < \frac{5}{2},

Decreasing

: \frac{5}{2} < x < \frac{11}{4},

Increasing

: \frac{11}{4} < x < \infty

Plug

x = \frac{11}{4}

into

\frac{5 e ^{4 x}}{2 x - 5} : 10 e^{11}

Minimum (\frac{11}{4}, 10 e^{11})

x^{3} - 6 x^{2} + 2 x + 4

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

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

f (x) = - 7 x^{3} + 21 x + 5

f (x) = 0.001 x^{2} + 2.4 x - 90

What is the extreme f(x)=(5e^{4x})/(2x-5) ?
The extreme f(x)=(5e^{4x})/(2x-5) is Minimum(11/4 ,10e^{11})