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.

$$\frac{2 + \sqrt{-4-4\cdot 1\cdot2}}{2}$$ Sorry for the poor layout, I don't really know how to lay a quadratic formula out on this site

  • $\begingroup$ Please let me know if I edited in a mistake with formatting... $\endgroup$ – apnorton Jan 8 '14 at 23:04
  • $\begingroup$ You need to find where the derivative is $0$. $\endgroup$ – André Nicolas Jan 8 '14 at 23:06
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    $\begingroup$ The stationary point is where the derivative is zero. Your square root expression looks like an attempt to compute a root of teh quaratic polynomial itself (and even for that the first $-4$ under the surd should be $(-2)^2=+4$) $\endgroup$ – Hagen von Eitzen Jan 8 '14 at 23:06
  • $\begingroup$ For classifying the point, it might be more useful to complete the square: x^2-2x+2 = (x +/- something)^2 +/- a constant. Then the nature of x=1 will be revealed. $\endgroup$ – Apprentice Queue Jan 8 '14 at 23:07

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.

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  • $\begingroup$ Thanks. Someone suggested then using quotient rule I don't know whether it's relevant, but anyway I looked it up.. I got 2x^3 - 4x^2 + 2x + 2 / (x^2 - 2x + 2)^2. I don't understand what i'm supposed to be doing with this new calculation.. $\endgroup$ – user119325 Jan 8 '14 at 23:25
  • $\begingroup$ @user119325: don't use the quotient rule for a non-quotient like $f(x)=x^2-2x+2$... $\endgroup$ – abiessu Jan 8 '14 at 23:28
  • $\begingroup$ A dude who's studying at Oxford told me, so I went with it.. =/ $\endgroup$ – user119325 Jan 8 '14 at 23:30

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:

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    $\begingroup$ Wrong, especially as the stationary point is $(1,1)$. $\endgroup$ – Hagen von Eitzen Jan 8 '14 at 23:09
  • $\begingroup$ This is good for classifying, but not good for the problem the OP is experiencing. $\endgroup$ – apnorton Jan 8 '14 at 23:10

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.

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  • $\begingroup$ "free form" algebraic manipulations, introduction of infinite intervals, intuitive reasoning, etc; These require a higher/deeper level of mathematical understanding/intuition. I am not criticizing (really), but don't you think "basic mechanics" and a basic understanding of why/how/when to use them, should be acquired before reaching out to the above? $\endgroup$ – user76568 Jan 9 '14 at 1:56
  • $\begingroup$ The fact that in intervals where the derivative is positive, the function is increasing is very basic. If one does not have a firm grasp of that, then one has little understanding of the derivative. Basic mechanics is of course necessary, or used to be when machines could not do it. But all too often mechanics is used to attempt to replace the basic understanding. $\endgroup$ – André Nicolas Jan 9 '14 at 2:10
  • $\begingroup$ As an analogy from Physics: You first learn Newton's mechanics (With the 3 "arbitrary" axioms/laws). Then, optionally, you study forms of analytical mechanics and, as a side effect, you are able to derive Newton's laws from mathematical reasoning. But, in QM, you may have your operators defined, a wave function "defined/collapsed", but how will you find the equation for the system if you did not learn classical mechaincs? $\endgroup$ – user76568 Jan 9 '14 at 2:10
  • $\begingroup$ I cannot comment on Physics. However, my discomfort with the second derivative test (in one-variable calculus) comes from years of experience grading calculus final exams. Students who attempt to use a second derivative test distressingly often get wrong answers. $\endgroup$ – André Nicolas Jan 9 '14 at 2:14
  • $\begingroup$ I see. I will not argue with statistical facts :). Regardless, from my experience as a Math/Physics student: When I find my self stuck solving an exam, and the distress kicks in, it is likely that I "find my way out" using "blind" mechanics and calculations (Which always yield all the potential points :), AND it is equally likely I find salvation by only reasoning, intuition, and "free form" thought (I get the points for that as well, strangely enough :). $\endgroup$ – user76568 Jan 9 '14 at 2:29

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