Calculus Part-One

Function and Inverses of Function

Function

A function is a relation between sets of objects that can be thought of as a “mathematical machine.”

This means for each input, there is exactly one output. Let’s say this explicitly.

Definition A function is a relation between sets, where for each input, there is exactly one output.

Moreover, whenever we talk about functions, we should try to explicitly state what type of things the inputs are and what type of things the outputs are. In calculus, functions often define a relation from (a subset of) the real numbers to (a subset of) the real numbers.

Warning

A function is a relation (such that for each input, there is exactly one output) between sets and should not be confused with either its formula or its plot.

• A formula merely describes the mapping using algebra.
• A plot merely describes the mapping using pictures.

Inverses of Function

If a function maps every input to exactly one output, an inverse of that function maps every “output” to exactly one “input.”

While this might sound somewhat esoteric, let’s see if we can ground this in some real-life contexts.

Example:

Suppose that you are filling a swimming pool using a garden hose—though because it rained last night, the pool starts with some water in it.

The volume of water in gallons after t hours of filling the pool is given by:
v(t) = 700t + 200
What does the inverse of this function tell you? What is the inverse of this function?

Solution

While v(t) tells you how many gallons of water are in the pool after a period of time, the inverse of v(t) tells you how much time must be spent to obtain a given volume.

To compute the inverse function, first set v = v(t) and write

[v = 700t + 200]

Now solve for :

[{t = } {v \over 700} − {2 \over 7}]

This is a function that maps volumes to times, and (t(v) = {v \over 700}−{2 \over 7}).

A Word on Notation

Given a function f (x), we have a way of writing an inverse of f (x), assuming it exists
(f^{-1}(x))

(f^{-1}(x)) = the inverse of f(x), if it exists.

On the other hand,

[f (x)^{-1} = 1 f (x)]

[f^{-1}(x) = {1 \over f(x)}]