### Differential and Integral Calculus (21-120) — Feedback on Homework 1

Homework 1 was due on Thursday 29th October 2013 and consisted of:

• Section 1.1 Q 64, 70, 76
• Section 1.3 Q 32, 36
• Section 1.5 Q 2, 14

The questions I marked were 1.1/76 (out of 3), 1.3/36 (out of 3) and 1.5/14 (out of 4).

Section 1.1 Q76. Common errors included:

• Not proving the result: it's important to communicate your reasoning! Simply stating a result isn't good enough for full credit.
• Not proving that $f(-x)=-f(x)$ for all values of $x$. I saw a lot of people check that $f(-x)=-f(x)$ for a few (or often just one) value of $x$. But you'll find that many functions, which are not odd, do have some values of $x$ for which $f(-x)=-f(x)$. As a really silly example, consider the function $f$ defined by $$f(x) = \begin{cases} 0 & \text{if}\ x \ne 2 \\ 1 & \text{if}\ x = 2 \end{cases}$$ This function is almost odd, since whenever $x \ne \pm 2$ we have $$f(-x) = 0 = -0 = -f(x)$$ and yet $f$ is not odd since $$f(-2) = 0 \ne -1 = -f(2)$$ Drawing a table with a few values of $x$ which shows that $f(-x)=-f(x)$ for those values is not sufficient for showing that $f$ is odd. However, finding a single value of $x$ for which $f(-x) \ne -f(x)$ is sufficient for showing that $f$ is not odd.
• Only proving that $f$ is not even. There are functions that are even, odd, both and neither. For instance:
• $f(x)=x^2$ is even but not odd
• $g(x)=x$ is odd but not even
• $h(x)=0$ is both odd and even
• $k(x)=x+1$ is neither odd nor even
Unlike integers, "not even" does not mean the same thing as "odd". You'll find situations like this all over the place: always work from the definitions, don't let your knowledge of the English language deceive you.
• Doing strange things with $|x|$. Remember the function $|x|$ is defined by $$|x| = \begin{cases} x & \text{if}\ x \ge 0 \\ -x & \text{if}\ x \lt 0 \end{cases}$$ If you give it a positive number, it just gives you your number back; if you give it a negative number, it flips its sign. For example, $|4|=4$ and $|{-3}|=3$. Some people wrote things like "$-x|x|=-x(x)=-x^2=\cdots$"; this is not correct!
• Using the result "odd × even = odd". This result is true, but if you wanted to use it you should have proved it, and proved that $x$ is odd and $|x|$ is even! In this case it was best just to prove directly that $f(-x)=-f(x)$.

Section 1.1 Q36. Common errors included:

• Most people correctly identified that $$(f \circ g)(x) = \frac{\sin 2x}{1+\sin 2x}$$ Most people then had the right idea in finding the domain: it's those values of $x$ for which $1+\sin 2x \ne 0$. What most people did here was, presumably, to use their knowledge of values of trig functions (or a calculator, or their computer) to identify that $\frac{3\pi}{4}$ was one such value, or $-\frac{\pi}{4}$ or whatever. So most of the almost-correct solutions I saw said something like "$x \ne \frac{3\pi}{4}$". However, this alone is not a correct answer. You need to identify all the values of $x$ for which $1+\sin 2x = 0$. This can be done using periodicity: for any value of $\theta$, $\sin(\theta) = \sin(\theta+2\pi)$. Thus: \begin{align} 1 + \sin 2x = 0 & \Leftrightarrow \sin 2x = -1 && \\ &\Leftrightarrow 2x = \frac{3\pi}{2} + 2k\pi && \text{for some integer k}\\ &\Leftrightarrow \ \,x = \frac{3\pi}{4} + k\pi && \text{for some integer k} \end{align} The solution I was looking for was $x=\frac{3\pi}{4}+k\pi$, or something equivalent (e.g. $x=-\frac{\pi}{4}+k\pi$).
• Another abundant error was in calculating the domain of $f \circ f$. Those who correctly simplifed $$(f \circ f)(x) = \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}} = \frac{\frac{x}{1+x}}{1+\frac{x}{1+x}} \cdot \frac{1+x}{1+x} = \frac{x}{1+2x}$$ then usually declared that the domain was $x \ne -\frac{1}{2}$. But let's look more closely at what we did when we simplified $f \circ f$: at some stage you multiply by $\frac{1+x}{1+x}$. This is not defined when $x = -1$, and so it cannot be in the domain of $f \circ f$. So the simplification given above is only valid when $x \ne -1$. Thus the domain of $f \circ f$ is all real values of $x$ except $-1$ and $-\frac{1}{2}$.
• A lot of people had trouble simplifying their expressions. Sometimes this led to the wrong answer. In these cases I gave partial credit for writing down $\frac{\frac{x}{1+x}}{1+\frac{x}{1+x}}$. If you're having any problem with simplifying complicated expressions, let me know and I'll see if we can arrange an extra class or something.
• Not simplifying answers. This time, I didn't deduct marks for this, but it's something you should get into the habit of. Leaving $$\frac{\frac{x}{1+x}}{1+\frac{x}{1+x}}$$ as a final answer isn't very illustrative. If you want to be able to do stuff with your mathematics then you need to present your working in a form that is usable!

Section 1.5 Q14. This was mostly done very well. The most common error was not indicating where the graph crosses the $y$-axis. In general you should always label where your graphs cross the axes, so long as it's reasonable to do so. In this case it was very easy to calculate, so it should have been labelled.

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