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I'm new here and I would like to know what teachers have saw in their experience about errors in students exams;

I'm interested to know what are the most common errors in exams about "calculus", more specific about:

  • studying a function with $\ln$, $e$, $x^2/x$, $\cos$, $\sin$... (domain, zeros, $f'(x)$, $f''(x)$, asymptote)
  • calculating an integral (indefinite and definite) by parts and substitution
  • calculating the derivative of complex/compound functions
  • theorems like Lagrange, Rolle and Bolzano...

I'm asking this because I'm preparing for an exam and I would like to know what are the most common errors, from the "smallest" one to the "biggest" one..

thanks for your help

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    $\begingroup$ Trouble with minus signs! $\endgroup$
    – André Nicolas
    Nov 14, 2011 at 13:53
  • $\begingroup$ May I ask to put it as answer in order to vote up? thanks $\endgroup$
    – Totty.js
    Nov 14, 2011 at 13:54
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    $\begingroup$ Spelling "asymptote"... ;) $\endgroup$
    – Hans Lundmark
    Nov 14, 2011 at 14:05
  • $\begingroup$ fixed, thanks... $\endgroup$
    – Totty.js
    Nov 14, 2011 at 14:14
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    $\begingroup$ You may already know about this (it was probably among the 10 most popular web pages in math for several years in the earlier days of the internet), but Eric Schechter's The Most Common Errors in Undergraduate Mathematics is a pretty good collection of student errors. $\endgroup$
    – Dave L. Renfro
    Nov 14, 2011 at 16:18

8 Answers 8

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Errors in routine algebra, in particular problems with minus signs, often turn a very "doable" problem into one that the student cannot do. If this happens early enough in a problem, even partial credit can disappear.

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The properties of logarithms and exponentials are often wrong. For example $\ln(xy)=\ln(x)+\ln(y)$, not $\ln(xy)=\ln(x)\ln(y)$. Not knowing $\sin$ and $\cos$ for the "special angles" like $0, \frac{\pi}{6}, \frac{7\pi}{4},$ etc.

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  • $\begingroup$ Or, they try to split up $\ln(x + y)$ as something or other. $\endgroup$
    – GeoffDS
    Nov 14, 2011 at 15:20
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Freshman's folly is all too common, $$(a+b)^n=a^n+b^n$$ for $|n|>1$. Although this does hold in finite fields...

My other two "favourites" are, $$\begin{align*}x^2-x&=0\\\Rightarrow x^2&=x\\\Rightarrow x&=1\end{align*}$$ and $$\begin{align*}x^2-4&=0\\\Rightarrow x^2&=4\\\Rightarrow x&=2\end{align*}$$

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    $\begingroup$ your other favorites are right, right? xD $\endgroup$
    – Totty.js
    Nov 14, 2011 at 14:31
  • $\begingroup$ I refuse to reply to such a comment! $\endgroup$
    – user1729
    Nov 14, 2011 at 14:33
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    $\begingroup$ @Totty: $x=0$ is also a solution to the first polynomial. The problem is that $x^2=x\Rightarrow x=1$ if $x\neq 0$, as otherwise you are dividing by zero. So the working should be followed by "$...x=1$ if $x\neq 0$", and then the student should realise that $x=0$ is also a solution. However, the frustrating thing is that they shouldn't be approaching problems like this at university! They should be taking out a common factor, so the first problem should be $x^2-x=0\Rightarrow x(x-1)=0\Rightarrow (x=1$ or $x=0)$. $\endgroup$
    – user1729
    Nov 14, 2011 at 14:41
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    $\begingroup$ @Totty: Simple: Never divide by a variable (if your polynomial is in $x$, do not divide by $x$. If it is in $x$ and $y$ do not divide by $x$ or $y$.). Instead, take out a common factor (as I pointed out above - $x^2-x=0\Rightarrow x(x-1)=0$. Similarly, the second problem should be $x^2-4=0\Rightarrow (x-2)(x+2)=0$ etc.). However, you can divide by a constant. For example, the solutions to $x^2+x-5=0$ are the same as the solutions to $2x^2+2x-10=0$. Can you see why this is? $\endgroup$
    – user1729
    Nov 14, 2011 at 14:47
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    $\begingroup$ Well, learn them. In the comments of a post isn't really the place to learn them. You are better asking someone in real life... $\endgroup$
    – user1729
    Nov 14, 2011 at 15:06
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  • After finding the critical points of $f$, many students forget to verify that they correspond to maxima/minima using second (or higher order) derivatives.

    This happens especially for the "word problems" where it is intuitively clear--from both the wording of the problem and physical intuition--that the critical point we get should correspond to a minimum/maximum. (Example problem: Find the dimensions of the cuboid with a given surface area $A$ such that the volume is maximized.)

  • It is possible that a function $f$ is increasing everywhere in an interval, but the derivative of the function is not strictly positive (just nonnegative). The standard example of such a function is $f(x) = x^3$ over $[-1, 1]$.

    From my experience, if students are asked to provide a counter-example for the fact, they are generally able to remember and produce one. On the other hand, if the question asks for the region where a given function is strictly increasing, then they are more likely to make a mistake out of haste.

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Often, students erroneously identify $(c, f'(c))$ as the critical point (or $(c, f''(c))$ as inflection point), after finding c to be the root of $f'(c) =0$ (or $f''(x) =0$ )

In this case, the point is $(c ,f(c))$.

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  • $\begingroup$ Of course, the first/second derivative tests might further confuse a student (who doesn't know where to "plug-in" c...) $\endgroup$
    – The Chaz 2.0
    Nov 14, 2011 at 14:13
  • $\begingroup$ yes that's good.. it's never explained in very detail when are done by teachers.. $\endgroup$
    – Totty.js
    Nov 14, 2011 at 15:21
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${1\over 3}+{2\over 5}={3\over 8}$.

And: "if $f'(a)>0$, then $f$ is increasing near $a$".

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  • $\begingroup$ What's wrong with the second one? By definition, for $ |h|$ sufficiently small, if $h\gt 0$, $f'(a)\gt 0$ means that $f(a+h)\gt f(a)$,or f is increasing on $(a,a+h)$. If $h\lt 0$ then $f(a+h)\lt f(a)$, or f is increasing on $(a+h,a)$. $\endgroup$
    – Divide1918
    Jun 25, 2020 at 7:47
  • $\begingroup$ @Divide1918 Increasing on $I$ means $f(y)> f(x)$ whenever both $x$ and $y$ are in $I$ with $y>x$. You only showed there is an $h$, as small as you like, with $f(a+h)>f(a)$ ($a$ is fixed here). See this for a counterexample of my statement. $\endgroup$
    – David Mitra
    Jun 25, 2020 at 8:41
  • $\begingroup$ So basically f has to be $C^1$ for that statement to hold? $\endgroup$
    – Divide1918
    Jun 25, 2020 at 9:55
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The following are not only common mistakes that students make, but are often glossed over.

1: For all $x$ and all $r,s$ $(x^r)^s=x^{rs}$

This is false since if we define $x=-1,r=2,s=\frac{1}{2}$ Then this "rule" gives us $$((-1)^2)^{\frac{1}{2}}=(-1)^1=-1.$$ On the other hand, $$(((-1)^2)^{\frac{1}{2}})=1^{\frac{1}{2}}=1.$$

2: Dividing ratios of functions and forgetting the domain. For instance consider the following computation, $$\frac{x}{x}=1.$$ This is false since we are stating that this is true for all $x$. However if $x=0$ we have a zero in the denominator and so the expression $\frac{x}{x}$ is undefined when $x=0$. What should be written is $$\frac{x}{x}=1, x\neq0.$$

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$${\cos ^{ - 1}}x \ne \frac{1}{{\cos x}}$$and $$\begin{array}{lcl}\cos \left( {10} \right) = - 0.839071529076 & \hspace{0.5in} & {\mbox{in radians}}\\ \cos \left( {10} \right) = 0.984807753012 & \hspace{0.5in} & {\mbox{in degrees}}\end{array}$$

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