# Graphing $\frac{x^2-x+1}{2(x-1)}$

I need to graph

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

So I reduced it to make the derivative easy:

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

which has roots $$x = 0, x = 2$$

But the derivative of the original function and the one I made are different. I can't see why. The two have the same roots, but I can't see what i'm doing wrong.

The concavity was easy to find by the derivative I made, but i don't know what's happening

• What makes you think $\dfrac 1 2 - \dfrac 1 {2(x-1)^2}$ is different from $\dfrac{(x-2)x}{2(x-1)^2}\text{ ?}$ They are the same. ${}\qquad{}$ – Michael Hardy Dec 16 '14 at 0:09
• @BrianM.Scott : Look again! – Michael Hardy Dec 16 '14 at 0:10
• @Cameron: Somehow I read $1-x$ instead of $x-1$. – Brian M. Scott Dec 16 '14 at 0:11
• @Brian: Been there! – Cameron Buie Dec 16 '14 at 0:12
• @MichaelHardy sorry, I though the second wolpram query was graphing the derivative, but is the original function, so i though they were different – Guerlando OCs Dec 16 '14 at 0:12

They are, in fact, the same expression. Your solution gives:

$$f'(x) = \frac{1}{2} - \frac{1}{2 \left ( x - 1 \right)^2 }$$

And the other one gives:

$$f'(x) = \frac{x\left(x - 2 \right)}{2(x - 1)^2}$$

First, one could notice the common denominator of $2(x - 1)^2$ seen in both expressions. This should stick out to you. Try rewriting the derivative on the top into a single fraction, and then compare the two results.

Ah, but they aren't different! They're just written differently. Try combining your derivative's terms over their least common denominator, then factor the numerator by noting that it is a difference of squares. You should get the same thing as Wolfram's derivative of $f(x).$

• I though they were different because I saw the graph of the second one, and it was graphing the original function, not the derivative. By the way, how would you analyze its signal? – Guerlando OCs Dec 16 '14 at 0:04
• What do you mean by "signal"? – Cameron Buie Dec 16 '14 at 0:05
• positive or negative – Guerlando OCs Dec 16 '14 at 0:10
• Aha! I would call that the "sign," rather than the "signal," but now I think I understand. So, if $f'(x_0)>0,$ what does that say about the behavior of $f$ at $x_0$? What about if $f'(x_0)<0$? (If the answer is "I don't know," then that's fine, but think about it, first. This will come up a lot.) – Cameron Buie Dec 16 '14 at 0:14
• If I think of it as the compact derivative of the first link, then it's only a case of analyzing the signal of the numerator, since the denominator is always positive. But it isn't clear how is the sign of the function in my derivative – Guerlando OCs Dec 16 '14 at 0:17

Find the common denominator of your denominator: $$\frac 12 - \frac 1{2(x-1)^2} = \dfrac{(x-1)^2 - 1}{2(x-1)^2} = \frac{x^2 - 2x}{2(x-1)^2} = \frac{(x-2)x}{2(x-1)^2}$$

Note that the derivative is zero when the numerator (not the denominator) is zero. In terms of graphing the function, the solution(s) to $f'(x) = 0$ will yield the values of $x$ where extrema exist (equilibrium points where the function is neither increasing nor decreasing.)

$$x(x-2) = 0\iff x=0\;\;\text{or}\;\; x = 2$$

Test each solution value to determine which is a maximum, which a minimum. That will help you in graphing the function.