What is the critical (e^{2x})/(x-3) ?

The critical (e^{2x})/(x-3) is x= 7/2

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critical (e^{2x})/(x-3)

critical points

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

x = \frac{7}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{7}{2}, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{e ^{2 x}}{x - 3} : x < 3 or x > 3

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

is not defined at

x = 3

, therefore:

x = \frac{7}{2}

f (x) = x e^{- 4 x}

f (x) = 6 x^{4} + 8 x^{3}

y = \frac{2 x ^{3} + 2}{x ^{2}}

f (x) = \frac{x ^{3} + 1}{x ^{2}}

f (x) = (x^{3} + 8)^{5}