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

The extreme f(x)=-(e^{2x})/(-2x-5) is Minimum(-2, 1/(e^4))

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

extreme points

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

Minimum (- 2, \frac{1}{e ^{4}})

Solution steps

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

Find intervals:

Decreasing

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

Decreasing

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

Increasing

: - 2 < x < \infty

Plug

x = - 2

into

- \frac{e ^{2 x}}{- 2 x - 5} : \frac{1}{e ^{4}}

Minimum (- 2, \frac{1}{e ^{4}})

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

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

g (x, y) = 3 x^{2} + y^{2} - 9

f (x) = 2500 x - 5 x^{2} + 10000

f (x) = (10 - 2 x) (10 - 2 x) (x)

What is the extreme f(x)=-(e^{2x})/(-2x-5) ?
The extreme f(x)=-(e^{2x})/(-2x-5) is Minimum(-2, 1/(e^4))