Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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 $x=1$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.