I'm attempting a past paper and I have been asked to compute the derivative for $(x^2-2x+2)$ and from this I calculated $2x-2$.
Once I completed this, I was then asked to find and classify the stationary point.
I usually use quadratic formulas to start this off, but for some reason I'm receiving a maths error. If someone can help me out, much appreciated.
$$\frac{2 + \sqrt{-4-4\cdot 1\cdot2}}{2}$$
Stationary points are found by setting the derivative equal to zero, and solving for $x$; i.e., solving $f'(x) = 0$.
In your case, $f'(x) = 2x - 2 = 0 \iff x = 1$.
You can then determine whether there is a maximum or minimum value at $x = 1$, using your favorite method.
IF the function is defined as (The OP requested this definition in a comment bellow): $$f(x)=f[\ln (g(x))]=\ln (x^2-2x+2)$$ The first-order derivative is: $$\frac{df(x)}{dx}=f'(x)=\frac{dg(x)}{dx}\frac{df(g)}{dg}=(2x-2)(\frac{1}{x^2-2x+2})=\frac{2x-2}{x^2-2x+2}$$
Stationary points are by definition: points for which the rate-of-change of the function (1st-order derivative) equals $0$. Makes sense when you imagine it as well. We follow this definition: $$f'(x)=\frac{2x-2}{x^2-2x+2}=0\implies x=1$$
Now, let's take the 2nd-order derivative: $$\frac{d^2f(x)}{dx^2}=\frac{df'(x)}{dx}=f''(x)=[\ln(x^2-2x+2)]''=(\frac{2x-2}{x^2-2x+2})'=\frac{2x(2-x)}{({x^2-2x+2})^2}$$
Meaning: the rate-of-change of the rate-of-change (Think of it as measure of how much the function is curving, and in which direction).
We already know there is a stationary point at $x=1$, so let's evaluate $f''(1)$: $$f''(1)=\frac{2(2-1)}{({1^2-2+2})^2}=2$$
From this we deduce that the rate-of-change is increasing at $x=1$, and therefore we have a global minimum at $x=1$.
Now, $f(1)=\ln(1)=0$, and thus $(1,0)$ is the global minimum on the graph.
Here is a plot of $f,f',f''$ colored blue,purple,gray , and a black point marking the stationary point, which in this case is a global minimum:
Introduction: It turns out from comments that OP's function is actually $$\ln(x^2-2x+2),$$ and the question asks for the critical point(s), and for a classification.
Note first of all that $x^2-2x+2=(x-1)^2+1$. In particular, $x^2-2x+2$ is always positive.
Note also that the derivative of $\ln(x^2-2x+2)$ is $$\frac{2x-2}{x^2-2x+2}.\tag{1}$$ The derivative is $0$ precisely if $2x-2=0$, that is, if $x=1$. Depending on the local definition of critical point, that means that the critical point is $1$, or that it is $(1,0)$.
Classification: Here we come to the motivation for my answering the question. Distressingly, answers to this and similar questions by the OP work with the second derivative. We can more simply use the first derivative.
As remarked earlier, the denominator in Expression (1) is always positive. Thus (1) is positive precisely if $2x-2$ is positive, and negative precisely if $2x-2$ is negative.
Note that $2x-2$ is negative if $x\lt 1$, and $2x-2$ is positive if $x\gt 1$.
It follows that our original function $\ln(x^2-2x+2)$ is decreasing in the interval $(-\infty,1]$, and increasing in the interval $[1,\infty)$. Thus $\ln(x^2-2x+2)$ attains a local (and absolute) minimum at $x=1$.
Remark: The second derivative test tends to be popular with students, since it promises a mechanical approach to the classification of critical points. Unfortunately, computation of the second derivative can be messy, and therefore subject to error. Often, the manipulation needed to find the critical points is enough to give us the sign of the first derivative in the intervals of interest. This is enough for the required classification.
