What is the critical f(x)=ln(2x)-(x^2)/5 ?

The critical f(x)=ln(2x)-(x^2)/5 is x=sqrt(5/2)

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critical f(x)=ln(2x)-(x^2)/5

critical points

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

x = \frac{5}{2}

Solution steps

ln (2 x) - \frac{x ^{2}}{5}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{5}{2}, x = \frac{5}{2}, x = 0

Identify critical points not in

f (x)

domain

Domain of

ln (2 x) - \frac{x ^{2}}{5} : x > 0

ln (2 x) - \frac{x ^{2}}{5}

is not defined at

x = - \frac{5}{2}, x = 0

, therefore:

x = \frac{5}{2}

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

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

f (x) = x^{3} e^{2 x}

f (x) = - x^{3} - 6 x^{2}

f (x) = \frac{( x ^{4} + 1 )}{x ^{2}}

What is the critical f(x)=ln(2x)-(x^2)/5 ?
The critical f(x)=ln(2x)-(x^2)/5 is x=sqrt(5/2)