Two functions are said to be [i]equal[/i] if they have the same set of inputs and each input is mapped to the same output. That is, the functions [math]f[/math] and [math]g[/math] are equal if their domains are the same and for every element [math]x[/math] in the common domain, [math]f\left(x\right)=g\left(x\right)[/math].[br][br]Note that we did not say the two functions are equal if their domains and their images are equal. Think about why this second definition is not strong enough. Can you think of two functions with the same domain and same image that are not equal?
Consider the two functions [math]f\left(x\right)=x^2[/math] and [math]g\left(x\right)=\left|x\right|[/math]. For both the domain is [math]\mathbb{R}[/math] and the image is all non-negative real numbers. However if we compare the way these two functions actually act on their domains we see they are not equal. For example [math]f\left(-2\right)=4[/math] but [math]g\left(-2\right)=2[/math].
With this understanding of equality in mind, let's return to parameterized plane curves. Remember, the curve itself is the IMAGE of the function [math]\vec{c}\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)[/math]. We just saw above that same domain and same image are not enough to guarantee equality between two functions. [br][br]Hit animate below to see two distinct (i.e. not-equal) functions whose image curves are the same.
In what ways are the two functions displayed above not equal?
Consider the input [math]t=\frac{\pi}{2}[/math]:[br][math]\vec{c}\left(\frac{\pi}{2}\right)=\left(0,1\right)[/math] while [math]\vec{d}\left(\frac{\pi}{2}\right)=\left(0,-1\right)[/math] hence these functions are acting differently on the same input and so are not equal.
It's important to remember that when we look at the plane curve associated with a path in [math]\mathbb{R}^2[/math] we are looking at the image of the function drawn in the codomain. To understand the function itself it's not sufficient to just look at the static image. We need to view the [i]way[/i] the image was formed. Animation helps us spot when the same image is generated by two distinct functions. If we were to analyze the functions [math]f\left(x\right)=x^2[/math] and [math]g\left(x\right)=\left|x\right|[/math] by animating the way they produce their images, we would have an animation like the one below.
Some questions to ponder. Think about all the knowledge you have of the two functions [math]f\left(x\right)=x^2[/math] and [math]g\left(x\right)=\left|x\right|[/math]. Is there any way to glean that information from the animation above? Specifically is it possible to look at the animation above and see that:[br]1. Both functions fail to be injective.[br]2. Both functions have an absolute minimum at [math]x=0[/math].[br]3. [math]x^2[/math] is differentiable everywhere, but [math]\left|x\right|[/math] fails to be differentiable at [math]x=0[/math].[br]4. The derivative of [math]x^2[/math] is a variable linear function while the derivative of [math]\left|x\right|[/math] (where it exists) is one of just two different constant functions.[br]5. [math]x^2[/math] is concave up across its domain, but [math]\left|x\right|[/math] has no concavity anywhere on its domain.