setting a function equal to another with the same variable

[credo]

In this problem speed is a function of distance
d = distance
f(d) = speed

now I need to create a second function that phase shifts the original to the left 1/2 mile (basically 1/2 mile of the first function will equal 0 of the second function). The issue I have is with defining the second function, it just doesn't seem right for whatever reason to use the same variable. This is really part of a bigger problem but it's nagging me and I can't continue:

g(d) = f(d - 1/2 mile)

Any one see a problem with this?



another, less clean way, would be:
d = distance
e = distance - 1/2 mi

f(d) = speed
g(e) = speed

but I really dont like that at all.

 

[kasie-tutor]

Dear credo,

g(d) = f(d - 1/2 mile)
Any one see a problem with this?

Here's the rule for a horizontal shift (translation) to the left, along with the other translation rules. The horizontal shifts are opposite of what is intuitive:

\(\displaystyle f(x + c)\) shifts \(\displaystyle f(x)\) to the LEFT \(\displaystyle c\) units.
\(\displaystyle f(x - c)\) shifts \(\displaystyle f(x)\) to the RIGHT \(\displaystyle c\) units.
\(\displaystyle f(x) + c\) shifts \(\displaystyle f(x)\) UP \(\displaystyle c\) units.
\(\displaystyle f(x) - c\) shifts \(\displaystyle f(x)\) DOWN \(\displaystyle c\) units.

That is the only problem I see. Your new equation \(\displaystyle g(d)\) is RIGHT-shifted of \(\displaystyle f(d)\).

Since the old equation and new equation are both functions of distance, it is right to use a "d" both times as the independent variable. Naming your new equation g(d) is good.

You don't need the "mile" unit in your equation.