# Finding critical numbers

I am having trouble finding critical numbers, specifically finding the roots or zeroes of a function. Especially when it involves a fraction.

For example right now I have \begin{align*} f(x) &= x^2 - x - \ln x\\ f' (x) &= 2x - 1 - \frac {1}{x}\\ f' (x) &= 2x - 1 - \frac {1}{x} = 0\\ f' (x) &= 2x - \frac {1}{x} = 1\\ f' (x) &= 2x^2 - 1 = x. \end{align*} Here I multiplied by $x$ on both sides which seems to be okay to do but it does not give me the same answer as if I were to plug in a number, for example 11, into \begin{align*} f' (x) &= 2x - 1 - \frac {1}{x} = 0\\ f' (x) &= 2x^2 -x - 1 = 0\\ \end{align*}

I know these are basic math concepts I should have mastered by now but I can't figure this out. Shouldn't both those function be equal to eachother?

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You're supposed to find the roots of $2x^2-x-1=0$; you could use either the quadratic formula or remember how to factor quadratic polynomials like this... – J. M. Mar 25 '12 at 1:42
Why is what I am doing wrong though? I did everything right according to math. – user138246 Mar 25 '12 at 1:45
Note: the third line in your list of equations is incorect. It is true that $2x-1-\frac{1}{x}=0$ is equivalent to $2x-\frac{1}{x}=1$, but $2x-\frac{1}{x}$ is not equal to $f'(x)$ anymore. – Arturo Magidin Mar 25 '12 at 3:15

You are fine down to $2x-1-\frac 1x=0$ and multiplying by $x$ is fine, too (you know $x \ne 0$ as it is not in the domain of the logarithm), but moving the $1$ to the other side was wasted effort. Now you have $2x^2-x-1=0$ which you can either factor as $(2x+1)(x-1)=0$ or use the quadratic formula to get the same result. As the domain of the logarithm is $x\gt 0$, the only critical point is $x=1$
Is it wrong to factor as $(2x+1)(x-1)$? – user138246 Mar 25 '12 at 2:01
@Jordan: not at all. You want the roots of the polynomial. As I said, you get the same answer either way, but need to exclude the root $x= \frac {-1}2$ as it is not in the domain of the original function. – Ross Millikan Mar 25 '12 at 2:19