Home / inverse functions calc 1 / 5. Inverse Functions - WebAssign

5. Inverse Functions - WebAssign - inverse functions calc 1

5. Inverse Functions - WebAssign-inverse functions calc 1

5. Inverse Functions
The inverse f -1 of a function f : A B "undoes" what f did.
That is, if f (x) = y, then f -1(y) = x, so f sends x to y, while f 1
sends y back to x. It goes without saying that this f -1 will only be
a function if f -1(y) is unambiguous, that is, when there is only one
x A so that f (x) = y. In that case, and only in that case, it is clear
that f -1(y) = x.
Let us now formalize these concepts.
Definition 1. A function f : A B is called one-to-one if it sends
different elements into different elements, that is, if x = x implies that
f (x) = f (x ).
One-to-one functions are also called injective functions or injections.
Visually, no horizontal line can intersect the graph of a one-to-one
function more than once.
For instance, if A and B are both the set of real numbers, then
f (x) = x and g(x) = x3 are both one-to-one, but h(x) = x2 is not.
Definition 2. Let f be a one-to-one function with domain A and
range B. Then the inverse of f is the function f -1 : B A given by
f -1(y) = x if f (x) = y.
Example 2. Let A and B both be the set of all real numbers. Let
f : A B be given by f (x) = 2x + 7. Then f -1(y) = (y - 7)/2.
Solution: If f (x) = y, then y = 2x + 7, so y - 7 = 2x and
(y - 7)/2 = x. As x = f -1(y), it follows that f -1(y) = (y - 7)/2.
The preceding example shows a general strategy for finding the in-
verse of a function. Write the equation f (x) = y, with the appropriate
algebraic expression replacing f (x). Then solve for x. If there is more
than one solution, then f is not one-to-one, and so it has no inverse
function. If there is one solution, then that expression is the value of
f -1(y).
Example 3. If A is the set of positive real numbers, B is the set
of real numbers that are larger than 1, and f : A B is given by
f (x) = x2 + 1, then f -1(y) = y - 1.
Solution: We have f (x) = x2 + 1 = y. So x2 = y - 1, and because we
know that x is positive and y > 1, we can take the square root of both
sides, leading to x = y - 1. Hence, f -1(y) = y - 1. 2
Finally, we point out that if f is a one-to-one function with domain
A and range B, then f -1 f is the identity function of A and f f -1
is the identity function of B.
For instance, using the functions of Example 3, for all positive real
numbers x, the identity (f -1 f )(x) = (x2 + 1) -1 = x2 = x
holds, and for all y > 1, the identity (f f -1)(y) = ( y - 1)2 + 1 =
y - 1 + 1 = y holds.
5.1. Logarithmic Functions. If a function contains only additions, sub-
tractions, multiplications, and divisions, then its inverse is often easy
to compute. Power functions, that is, functions of the form f (x) = x,
where is a real number, are not much more difficult. However, what
is the inverse of an exponential function?
Let f (x) = 2x. It is easy to see, by plotting the graph of f or
otherwise, that f is a one-to-one function whose domain is the set of
all real numbers and whose range is the set of all positive real numbers.
So the inverse of f is a function from the set of positive reals to the set
of all reals. But what is that inverse function f -1? By the definition
of inverse functions in general, this is the function that sends 2x to x
for all positive real numbers 2x. In particular, f -1(2) = 1, f -1(4) = 2,
f -1(32) = 5, and f -1(1/2) = -1. That is, f -1(y) tells us to what power
we have to raise 2 if the result is to be y. This important concept has
its own name.
Definition 3. Let m be a positive real number. Then the inverse
of the function f (x) = mx is called the logarithmic function with base
m, and is denoted by logm.
So if f (x) = xm = y, then logm(y) = x. For instance, log2(64) = 6,
log3(81) = 4, log5(1/25) = -2, and log0.5(16) = -4.
Logarithmic functions satisfy certain rules that are very similar to
those satisfied by exponential functions and can, in fact, be deduced
from them. These are
(I) log(xy) = log x + log y.
(II) log(x/y) = log x - log y.
(III) log (xa) = a log x.
(IV) log b x = log x .
(V) aloga x = x.
(VI) loga(ax) = x.
The last two rules simply express the fact that the functions f (x) =
ax and f -1(y) = loga(y) are inverses of each other, so their composition
is an identity function.
If we know the logarithm of a number in a base and want to compute
it in another base, we can do so using the following theorem.
Theorem 1. For positive real numbers a, b, and x, we have
loga x = logb(x) .
Proof. Start with the identity
x = aloga x.
Now take the logarithm of base b of both sides to get
logb x = loga logb ax.
Now divide both sides by logb to get the identity of the theorem. 2
Example 4. We can use Theorem 1 to compute log16(256) from
log2(256) as follows:
log16(256) = log2(256) = 8 = 2.
log2(16) 4
So if a calculator or computer can provide the logarithm of all
positive real numbers in one base, we can compute the logarithm of any
positive real number in any base. For this reason, many calculators and
computers are programmed to work primarily with logarithms of one
given base, namely of base e, where e 2.718 is an irrational number
that will be formally defined in Chapter 2.
The logarithm of base e is so important that it has its own name,
natural logarithm, and its own notation, ln. So ln x = loge x.
5.2. Inverses of Trigonometric Functions. Basic trigonometric functions,
such as sin, cos, and tan, are very important in calculus, so it is no sur-
prise that their inverse functions are important as well. However, we
have to be precise when we define them since trigonometric functions
are not one-to-one. In fact, they are periodical, of period 2 or , and
so they take every value in their range infinitely often.
In order to get around this difficulty, we will restrict our trigono-
metric functions to just a short interval, in which they are one-to-one,
and define their inverses based on that restriction.
For instance, consider sin as a function whose domain is [-/2, /2].
In that interval, sin is a one-to-one function (since it is increasing),
and its range is the interval [-1, 1]. So its inverse is the function

How to find the formula for an inverse function? The inverse f -1 (x) = 3 + sqrt [ (x+5)/2] Start with x: x Add 5: x+5 Divide by 2: (x+5)/2 Take the square root: ± sqrt [ (x+5)/2] Add 3: 3 ± sqrt [ (x+5)/2] Wait! That inverse isn't a function because there are two values of y for every x. ...