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

The critical f(x)=x^2-x-ln(x) is x=1

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

critical points

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

x = 1

Solution steps

x^{2} - x - ln (x)

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{2}, x = 1, x = 0

Identify critical points not in

f (x)

domain

Domain of

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

x^{2} - x - ln (x)

is not defined at

x = - \frac{1}{2}, x = 0

, therefore:

x = 1

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

f (x, y) = e^{(x^{2} + 0.5 y^{2} - 5 x y - 3 x)}

65 x^{\frac{6}{7}} + x^{\frac{13}{7}}

\frac{4 + x}{x - 4}