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
 A: 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 $f(x).$
A: 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.
A: 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.
