What is the extreme f(x)=(2e^{3x})/(-2x-5) ?

The extreme f(x)=(2e^{3x})/(-2x-5) is Maximum(-13/6 ,-3/(e^{13/2)})

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extreme f(x)=(2e^{3x})/(-2x-5)

extreme

Maximum (- \frac{13}{6}, - \frac{3}{e ^{\frac{13}{2}}})

Solution steps

f^{'} (x) = \frac{2 ( - 6 e ^{3 x} x - 13 e ^{3 x} )}{( - 2 x - 5 ) ^{2}}

Find intervals:

Increasing

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

Increasing

: - \frac{5}{2} < x < - \frac{13}{6},

Decreasing

: - \frac{13}{6} < x < \infty

Plug

x = - \frac{13}{6}

into

\frac{2 e ^{3 x}}{- 2 x - 5} : - \frac{3}{e ^{\frac{13}{2}}}

Maximum (- \frac{13}{6}, - \frac{3}{e ^{\frac{13}{2}}})

What is the extreme f(x)=(2e^{3x})/(-2x-5) ?
The extreme f(x)=(2e^{3x})/(-2x-5) is Maximum(-13/6 ,-3/(e^{13/2)})