Solutions to Homework 6
1. A function f (x) is called one-to-one if it satisfies the the following
property:
f (a) = f (b) =⇒ a = b.
Which of the following functions is one-to-one? If the given function
is not one-to-one, then given a specific example of two distinct real
numbers a and b in the domain for which f (a) = f (b).
(a) f (x) = 2x
This function is one-to-one. One way to see this is to determine
that it has an inverse function. Let y = 2x . We can solve for
x as follows: ln y = x ln 2 =⇒ x = (ln y)/(ln 2). Therefore,
f −1 (x) = (ln y)/(ln 2).
In this solution, we used the following property of logarithms:
ln ax = x ln a. Another solution is to use the base two logarithm:
f −1 (x) = log2 x.
Still another point of view is to use the definition of one-to-one
directly. If 2x = 2y , then by taking the logarithm of both sides,
we deduce that x ln 2 = y ln 2. Dividing by ln 2, we deduce that
x = y. Hence 2x = 2y =⇒ x = y.
(b) f (x) = x3
√
This function is one-to-one since it has as inverse f −1 (x) = 3 x.
Another point if of view is to use the definition of one-to-one
directly. If x3 = y 3 , then by taking the cube root (which is defined
for all real numbers, including negative numbers), we deduce that
x = y.
(c) f (x) = x3 − x
This function is not one-to-one. For example, f (0) = 0. But also
f (1) = 0. Of course 0 6= 1. So, f (x) is not one-to-one.
(d) f (x) = [[x]] − x, where [[x]] is the greatest integer less than or
equal to x.
This function is not one-to-one. For example, f (0) = 0 and also
f (1) = 0. (In fact, if n is an integer, then f (n) = 0.)
2. If f (x) is a one-to-one function of a real variable x, then its inverse
function can be found as follows: let y = f (x) and solve the equation
for x; this yields an equation x = g(y). The function g(x) is the inverse
of f (x).
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Here is an example. If f (x) = 2x + 7, then f (x) is one-to-one. To find
the inverse, we solve the equation y = 2x + 7 for the variable x. We
obtain x = (y − 7)/2. Therefore the function g(x) = (x − 7)/2 is the
inverse of f (x).
Find the inverse of each of the following functions.
(a) f (x) = x3 − 8
The inverse function is found by solving the equation y = x3 − 8
√
3
for
x.
We
determine
that
x
=
y + 8. Therefore, f −1 (x) =
√
3
x + 8.
(b) f (x) = e2x
Solve y = e2x for x by first taking the logarithm of both sides.
Since ln x is the inverse of ex , we can use the fact that ln ex = x.
Therefore, if y = e2x , then ln y = 2x. Hence x = (ln y)/2. And
so f −1 (x) = (ln x)/2.
(c) f (x) = 3 sin x, where −π/2 ≤ x ≤ π/2
To solve y = 3 sin x we use the arcsine function, which I will
denote by sin−1 x. You must first divide both sides of the equation
by 3. The reason why is that the arcsine function has the property
that sin−1 (sin x) = x; this property does not suggest in any way
that the term sin x can be preceded by a number (or anything
else). So, dividing by 3, we have that (y/3) = sin x; now taking
the arcsine of both sides, we have that sin−1 (y/3) = x. Therefore,
f −1 (x) = sin−1 (x/3). You may ask what is the domain of f −1 (x).
The answer is that the domain of f −1 (x) is equal to the range of
f (x). Since the range of f (x) = 3 sin x is [−3, 3], the domain of
f −1 (x) = sin−1 (x/3) is [−3, 3].
(d) f (x) = tan−1 (x/2)
To solve y = tan−1 (x/2), we apply the tangent function to both
sides of the equation to obtain tan y = x/2. Therefore x = 2 tan y.
So, f −1 (x) = 2 tan x, however we must restrict the domain to
(−π/2, π/2).
The domain and range of these functions can be determined as
in the previous problem. Since the range of f −1 (x) = 2 tan x is
equal to (−∞, ∞), the domain of f (x) is equal to (−∞, ∞). And
since the range of f (x) = tan−1 x/2 is (−π/2, π/2), the domain
of f −1 (x) is (−π/2, π/2). This is a subtle point. The issue is that
the tangent function is defined for all real numbers except those
2
numbers of the form π2 + nπ, where n is an integer. As given, the
tangent function is not one-to-one. However, if we restrict the
domain to (−π/2, π/2), the function becomes one-to-one. (This
is analogous to choosing either the positive square root or the
negative square root when solving y = x2 for x; a choice must
be made if you are to arrive at a function.) These matters are
discussed in more detail in section 1.5.
3. If f (x) and g(x) are inverses of one another, then their graphs are
mirror reflections of one another across the line y = x. Illustrate this
fact by sketching the graphs y = f (x) and the graph y = f −1 (x) for
each of the functions in problem 2 above. List the exact coordinates of
at least three points on each graph. You should use a graphing utility
such as fooplot.com or a calculator to assist with your sketches.
In the graphs below, there is a sketch of y = f (x), y = f −1 (x), and
the line y = x (to illustrate the mirror reflection).
(a) f (x) = x3 − 8, f −1 (x) =
√
3
x+8
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(b) f (x) = e2x , f −1 = (ln x)/2.
(c) f (x) = 3 sin x, f −1 = sin−1 (x/3).
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(d) f (x) = tan−1 (x/2), f −1 = 2 tan x.
4. The derivative of a function and its inverse are reciprocals of one another provided you evaluate the derivative at values which correspond
to each other via the function. Illustrate this by computing f 0 (2) and
(f −1 )0 (32) for the function f (x) = x5 .
√
First, let’s compute f −1 (x). Solving y = x5 for x, we obtain x = 5 y.
√
Therefore, f −1 (x) = 5 x = x1/5 .
Now we compute the derivatives: f 0 (x) = 5x4 and (f −1 )0 (x) = (1/5)x−4/5 .
Evaluating, we find that f 0 (2) = 5(2)4 = 80 and that (f −1 )0 (32) =
(1/5)(32)−4/5 = (1/5)(2)−4 = (1/5)(1/16) = 1/80.
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