What is the critical y=(xe^x-9x)/(e^x) ?

The critical y=(xe^x-9x)/(e^x) is x=-\W_{0}(e/9)+1

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critical y=(xe^x-9x)/(e^x)

critical points

y = \frac{x e ^{x} - 9 x}{e ^{x}}

x = - W_{0} (\frac{e}{9}) + 1

Solution steps

\frac{x e ^{x} - 9 x}{e ^{x}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - W_{0} (\frac{e}{9}) + 1

Identify critical points not in

f (x)

domain

Domain of

\frac{x e ^{x} - 9 x}{e ^{x}} : - \infty < x < \infty

All critical points are in domain

x = - W_{0} (\frac{e}{9}) + 1

x^{8} (x - 3)^{7}

f (x) = \frac{5 ( x + 4 )}{x ^{2} + x - 12}

f (x) = x - 9

ln (7 - 5 x^{2})

f (x) = 3 cos (x + \frac{π}{2})

What is the critical y=(xe^x-9x)/(e^x) ?
The critical y=(xe^x-9x)/(e^x) is x=-\W_{0}(e/9)+1