Question

Hello,

I am currently going through Problem set 1, I am at a loss of how to determine whether a function is odd or even and I'm pretty sure it was never alluded to in the lectures.

Could somebody take me through 1A-3a and 3b and hopefully I'll be fine with the rest.
Thank you.


http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1sol.pdf

Answer 1

It is considered a college algebra level skill.
In a given function replace every instance of x with -x.
If through simplification, the original function is returned, it is even.
If the exact opposite (negative) is returned, the function is odd.
If neither can be found... it is "neither even or odd"

Answer 2

The presented file has a few steps of the solutions, but I will try to break it down a bit more.
It looks like the first is probably:
\[f(x)=\frac{x ^{3}+3x}{1-x ^{4}}\]Put -x in for every x (I use parentheses for every substitution).\[f(-x)=\frac{(-x) ^{3}+3(-x)}{1-(-x)^{4}}=\frac{-x ^{3}-3x}{1-x ^{4}}\]If it doesn't look like the original f(x), I try to factor a negative to the front to see if I can make it really obvious that it is the opposite of f(x).\[f(-x)=-\left( \frac{x ^{3}+3x}{1-x ^{4}}\right)\]So f(-x) = -f(x) which means that f(x) is odd.

Answer 3

One reason that a person does this in calculus is to aid in graphing a function. Even and odd functions have symmetry, meaning that they have a mirror reflection. If you can graph one side, the other side can be assumed due to its symmetry.

The graph of an even function is symmetric across the y-axis. Look at the graph of f(x)=x^2. What occurs on one side of the y-axis is reflected, left versus right, to the other side.

The graph of an odd function is symmetric across the origin. Look at the graph of f(x)=x^3. What occurs on one side of the origin is reflected diagonally to the other side.

Given:\[f(x)=\sin x\]
You can see that there is symmetry across the origin, so it becomes easier to remember the identity sin(-x) = -sin(x) because of this even/odd stuff (math term ;) )
In your (b) example, squaring the function returns the original. Therefore it is even.\[f(x)=(\sin x)^{2}\]\[f(-x)=(\sin(-x))^{2}=(-\sin x)^{2}=(\sin x)^{2}\]