Example 2.3 Let f : R→ R and g : R → R be the functions f(x) = x2 and g(x) = 2x− 1. What are the functions g ◦ f and f ◦ g?
Here, as the functions both go from R to R, we can find both of these compositions.
In particular,
g◦ f is the function where
(g◦ f)(x) = g(f(x)) = g(x2) = 2x2− 1, where (g◦ f) : R → R.
f ◦ g is the function where
(f ◦ g)(x) = f(g(x)) = f(2x − 1) = (2x − 1)2, and (f ◦ g) : R → R.
Indeed, observe that as (2x− 1)2 = 4x2− 4x + 1, these are certainly not the same function.
Activity 2.5 Let f : R→ R and g : R → R be the functions f(x) = x2+ 1 and g(x) = 2x. What are the functions g◦ f and f ◦ g?
In particular, we will also need to be able to identify compositions the ‘other way’ when we cover the chain rule in Section 3.2.2. For instance, it should be clear that the
function (x2+ 5)3 is the composition of the function x3 after the function x2+ 5.
Activity 2.6 Explain why the function (x2+ 5)3 is the composition of the function x3 after the function x2+ 5.
2.1.3 Inverse functions
If A and B are sets and we have a function f : A→ B, we know that this means that for every x∈ A there is a unique y ∈ B such that y = f(x). Now, if we can define another function g : B → A, i.e. for every y ∈ B there is a unique x ∈ A such that
y = f (x) if and only if x = g(y),
then we call the function, g, the inverse of f and denote it by f−1. In terms of ‘black boxes’, this means that we have
x∈ A −→ f −→ f(x) ∈ B, for f and, if it exists, we have
y∈ B −→ f−1−→ f−1(y)∈ A, for f−1, or more usefully,
f (x)∈ B −→ f−1 −→ x ∈ A.
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2.1. Introduction: What is a function?
In particular, this means that if the inverse, f−1, of f exists, we see that the composition f−1 after f gives us
x∈ A −→ f −→ f(x) ∈ B −→ f−1−→ x ∈ A,
and so (f−1◦ f)(x) = f−1(f (x)) = x whereas the composition f after f−1 gives us y∈ B −→ f−1−→ f−1(y)∈ A −→ f −→ y ∈ B,
and so (f ◦ f−1)(y) = f (f−1(y)) = y. That is, the inverse of a function (if it exists)
‘undoes’ what the function does and vice versa.
The question, then, is how can we tell whether an inverse function exists? And, if it does exist, how can we find it? Well, given the function f : A→ B, the inverse will exist if we are able to take y = f (x) and solve it to obtain a unique solution, x, in terms of y for every y ∈ B. And, if we can do this, these solutions will tell us what the inverse function is, i.e. they will allow us to identify the function, f−1(y), by comparison with x = f−1(y). To make this clear, let’s look at an example.
Example 2.4 Consider the function f : R→ R given by f(x) = x + 2. Explain why this function has an inverse and find it.
Using the graph or common sense, we see that the function f (x) = x + 2 has an inverse, since every y ∈ R where y = f(x) gives rise to a unique x ∈ R given by x = y− 2. As such, we can conclude that the inverse of this function exists and we have x = f−1(y) = y− 2. Of course, we can now write this inverse as f−1(x) = x− 2 if we want it in terms of x.
Indeed, notice that, if we have the function f (x) and its inverse function f−1(x), the graph of f−1 is the reflection of the graph of f about the line y = x. This happens because any point (x, y) on the curve y = f (x) becomes, under a reflection about the line y = x, a point (y, x) on the curve x = f (y) which is the same as saying that y = f−1(x)!
Activity 2.7 Verify that the curve y = f−1(x) is the reflection about the line y = x of the curve y = f (x) using the function we saw in Example 2.4.
Of course, not every function has an inverse as the next example shows.
Example 2.5 Consider the function f : R→ R given by f(x) = x2. Explain why this function does not have an inverse.
If we take any y ∈ R where y = f(x) this gives us the equation y = x2 and, if we are considering x∈ R, this gives rise to a problem as far as the inverse of f is concerned because:
If y < 0, we get no solution for x as we know that x ∈ R means that y = x2 ≥ 0.
If y > 0, we get two solutions for x as we know that we can get x =±√y∈ R.
That is, we can find no inverse in this case since we cannot guarantee a unique solution for x∈ R from the equation y = x2 for all y ∈ R.
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Of course, we can usually get around such problems if we are prepared to restrict the domain and the co-domain of the function. But, in that case, we would be finding the appropriate local inverses as opposed to its inverse (which, remember, doesn’t exist!).
Activity 2.8 By considering the domains (−∞, 0] and [0, ∞) and suitably restricting the co-domain of the function in Example 2.5, find its local inverses.
Let’s now look at the inverses of the elementary functions we considered in Section 2.1.1.
Power functions: root functions
If we have the power function f (x) = xn where x∈ N and f : [0, ∞) → [0, ∞) we can see that the inverse is given by
x = f−1(y) = y1/n, and this is called a root function. Thus, we have
x = y1/n if and only if y = xn,
provided that x, y≥ 0. In particular, if n = 2, this is the square root function, i.e. we have y1/2= √y.
Activity 2.9 Draw the graph of the power function f : [0,∞) → [0, ∞) where f (x) = x2 and its inverse.
This also works for f (x) = xn where f : R→ R if n is odd. But, if n is even, the function f (x) = xn where f : R→ R does not have an inverse as we saw, for n = 2, in Example 2.5.
Activity 2.10 Explain why we can find an inverse of the function f : R→ R where f (x) = xn if n is odd. Why doesn’t this work if n is even?
Exponential functions: logarithmic functions
If we have the exponential function f (x) = ax where f : R→ (0, ∞) and a 6= 1 is a positive real number, the inverse is the function f−1: (0,∞) → R given by
x = f−1(y) = logay, which is the logarithm to base a. Thus, we have
x = logay if and only if y = ax, provided that y > 0.
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2.1. Introduction: What is a function?
Activity 2.11 Draw the graph of the exponential function f : R→ (0, ∞) where f (x) = 2x and its inverse, f−1(x) = log2x where f−1 : (0,∞) → R.
In particular, we see from this that as
(f ◦ f−1)(x) = f (f−1(x)) = x we have alogax = x, and as
(f−1◦ f)(x) = f−1(f (x)) = x we have logaax = x.
These results will be useful in Section 2.1.4 when we consider the laws of of logarithms.
Trigonometric functions: inverse trigonometric functions
If we want to discuss the inverses of the trigonometric functions sine and cosine, it is first necessary to restrict their domain due to their oscillatory nature. To do this, we consider a certain interval of values of θ, called the principal range, so that each value of the function corresponds to a unique value of θ. Indeed, for the:
sine function, we take the principal range to be the interval [−π2,π2] so that the function sin : [−π2,π2]→ [−1, 1] where y = sin θ has an inverse. This inverse is denoted by sin−1 (or arcsin) where sin−1 : [−1, 1] → [−π2,π2]. Thus, we have
y = sin θ if and only if θ = sin−1y, provided that −π2 ≤ θ ≤ π2 and −1 ≤ y ≤ 1.
cosine function, we take the principal range to be the interval [0, π] so that the function cos : [0, π]→ [−1, 1] where y = cos θ has an inverse. This inverse is denoted by cos−1 (or arccos) where cos−1 : [−1, 1] → [0, π]. Thus, we have
y = cos θ if and only if θ = cos−1y, provided that 0≤ θ ≤ π and −1 ≤ y ≤ 1.
It will also be convenient for us to consider the inverse of the tangent function where, as well as the oscillations, we need to take care to avoid the asymptotes that occur when this function is undefined. As such, for the
tangent function, we take the principal range to be the interval (−π2,π2) so that the function tan : (−π2,π2)→ R where y = tan θ has an inverse. This inverse is denoted by tan−1 (or arctan) where tan−1: R→ (−π2,π2). Thus, we have
y = tan θ if and only if θ = tan−1y, provided that −π2 < θ < π2.
In particular, observe that sin−1, cos−1 and tan−1 are the inverses of the functions sin, cos and tan respectively and not their reciprocals which we denoted by cosec, sec and cot respectively in Section 2.1.2!
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Activity 2.12 Find the acute angles θ1, θ2 and θ3 where θ1 = sin−1 12, θ2 = cos−1 12 and t3 = tan−11.
Also find cosec θ1, sec θ2 and cot θ3.
2.1.4 Identities
An expression such as
(x + 1)2 = x2 + 2x + 1,
which is true for all x is called an identity and, as you know, these are useful when we need to simplify expressions. In particular, in Chapter 1 of 173 Algebra, you saw that the power laws dictate that
aman= am+n, am
an = am−n and (am)n= amn,
and these are identities that work for any values of a, m and n for which both sides are defined. Indeed, these ‘laws’ allow us to simplify expressions that may result from appropriate products, quotients and compositions of power functions or exponential functions.
Activity 2.13 If f (x) = x3, g(x) = x4 and h(x) = 2x, find the functions (f · g)(x), (f /g)(x) and (g◦ h)(x) simplifying your answers as far as possible.
We now look at some other identities that will be useful in this course.
The laws of logarithms
For any positive real number a6= 1, the laws of logarithms state that logax + logay = loga(xy), logax− logay = loga
x y
and y logax = loga(xy), provided that all of the terms involved are defined. As you may know, these ‘laws’ are easily derived from the power laws we saw above and the fact that
alogax= x, which we saw earlier in Section 2.1.3.
Activity 2.14 Derive the laws of logarithms from the power laws.
It is also useful to note that if a, b6= 1 are positive real numbers, then we have the
‘change of base formula’ which states that
logax = logbx logba,
and this allows us to write logarithms to base a in terms of logarithms to base b.
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2.1. Introduction: What is a function?
Activity 2.15 Derive the change of base formula for logarithms.
Trigonometric identities
There are also identities that allow us to simplify various expressions involving the trigonometric functions. For instance, using the triangle in Figure 2.4, Pythagoras’
theorem allows us to see that sin2θ + cos2θ =
opposite hypotenuse
2
+
adjacent hypotenuse
2
= opposite2+ adjacent2 hypotenuse2
= hypotenuse2 hypotenuse2
= 1,
and so, for acute angles,2 we have shown that
sin2θ + cos2θ = 1. (2.2)
In particular, for natural numbers n≥ 2, note that we commonly abbreviate things like (sin θ)n by writing them as sinnθ. Further, dividing both sides of this expression by sin2θ we get
1 + cot2θ = cosec2θ, (2.3)
and this works as long as θ6= nπ for n ∈ Z whereas dividing both sides of this expression by cos2θ we get
tan2θ + 1 = sec2θ, (2.4)
and this works as long as θ6= (2n + 1)π2 for n∈ Z. We call these three identities the Pythagorean identities as they are simple consequences of Pythagoras’s theorem.
Activity 2.16 Use (2.2) to derive the Pythagorean identities (2.3) and (2.4).
Another useful pair of trigonometric identities are the compound-angle formulae given by
sin(θ + ϕ) = sin θ cos ϕ + cos θ sin ϕ and cos(θ + ϕ) = cos θ cos ϕ− sin θ sin ϕ, which work for all θ, ϕ∈ R.
Activity 2.17 Observe from the graphs of the sine and cosine functions in
Figure 2.7 that sine is an odd function, i.e. sin(−θ) = − sin θ, and cosine is an even function, i.e. cos(−θ) = cos θ. Use these facts and the compound-angle formulae to show that we also have
sin(θ− ϕ) = sin θ cos ϕ − cos θ sin ϕ and cos(θ − ϕ) = cos θ cos ϕ + sin θ sin ϕ, for θ, ϕ∈ R.
2Of course, if we consider how we extend the definitions of the sine and cosine functions to all θ∈ R, it should be clear that this identity is actually true for all θ∈ R.