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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

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  • $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 16 '14 at 0:09
  • $\begingroup$ @BrianM.Scott : Look again! $\endgroup$ – Michael Hardy Dec 16 '14 at 0:10
  • $\begingroup$ @Cameron: Somehow I read $1-x$ instead of $x-1$. $\endgroup$ – Brian M. Scott Dec 16 '14 at 0:11
  • $\begingroup$ @Brian: Been there! $\endgroup$ – Cameron Buie Dec 16 '14 at 0:12
  • $\begingroup$ @MichaelHardy sorry, I though the second wolpram query was graphing the derivative, but is the original function, so i though they were different $\endgroup$ – Guerlando OCs Dec 16 '14 at 0:12
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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.

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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 $f(x).$

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    $\begingroup$ 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? $\endgroup$ – Guerlando OCs Dec 16 '14 at 0:04
  • $\begingroup$ What do you mean by "signal"? $\endgroup$ – Cameron Buie Dec 16 '14 at 0:05
  • $\begingroup$ positive or negative $\endgroup$ – Guerlando OCs Dec 16 '14 at 0:10
  • $\begingroup$ 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.) $\endgroup$ – Cameron Buie Dec 16 '14 at 0:14
  • $\begingroup$ 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 $\endgroup$ – Guerlando OCs Dec 16 '14 at 0:17
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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.

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