Inverse Functions

An inverse function of any given function will undo the given function. If f(x) is the given function and g(x) is its inverse, then (gof)(x) = x. We will explain this aspect a bit more clearly later. Another useful definition we need to remember is that g(x) satisfies the reverse ordered pairs that satisfy f(x).

In such a case, it is algebraically written as,  g(x) = f-1(x). It is to be noted that the -1 above refers to the inverse of the function f(x) and not the reciprocal of the function f(x). To avoid confusion, some authors prefer to use the term arc to denote an inverse. That is, the inverse of the function f(x) is denoted as arc f(x). However, the first representation is more common and hence we will use that form of representation for inverse functions in this article.

Inverse Trig Functions

The most common inverse functions are inverse trigonometric functions. A normal trigonometric function gives values for different input angles. Inverse trigonometric functions, on the other hand give us the angles for known values. The importance of the inverse trigonometric function is explained with the following example:

We know that tan 30° = $\frac{(\sqrt{3})}{3}$. That is, the slope of line is $\frac{(\sqrt{3})}{3}$. In certain statutory regulations, the slopes of some constructions are not to exceed $\frac{(\sqrt{3})}{3}$. How does the builder determine the angle for such constructions? It becomes very simple if he knows that inverse $\tan^{-1}$$\frac{\sqrt{3}}{3}$ =30°

Thus, in general, the domain and the range of trigonometric functions get reversed when they become inverse trigonometric functions.

Find the Inverse of a Function


The method of finding the inverse of a function is explained below:

Suppose f(x) is the given function. Now f(x) = some expression with the variable x.

Assume, y = the same expression with the variable x.

In other words, the same expression with the variable x = y

Now solve for x in terms of y. If this is not possible, then the given function is invertible, meaning it has no inverse.

If the above solution is possible, then switch the variables. That is, now the solution is y in terms of x, which is the inverse of the original function.

Let us illustrate the above method with an example.

If f(x) = 2x + 1, then what is f-1(x)?

Let, y = 2x + 1

or, 2x + 1 = y
or, 2x = y -1
or, x = $\frac{(y – 1)}{2}$

Switching the variables, y = $\frac{(x – 1)}{2}$

Which means, f-1(x) = $\frac{(x – 1)}{2}$

Inverse Function Theorem

There are two simple theorems related to inverse functions.

Let f(x) and g(x) are two functions of a single variable. If (fog)(x) and (gof)(x) are both the same and if both are equal to x, then f(x) and g(x) are inverse functions.

Let us take the same example where (x) = 2x and g(x) = $\frac{x}{2}$.

(fog)(x) = 2($\frac{x}{2}$) = x and (gof)(x) = $\frac{2x}{2}$ = x . Hence as already shown, f(x) and g(x) are inverse functions.

The second inverse function theorem is based on the derivative of functions. If f(x) and g(x) are continuous functions and differentiable at a point a, and if,

g' (b) = $\frac{1}{f'(a)}, f'(a) $\neq$ 0, b= f(a) then f(x) and g(x) are inverse functions.

Let us verify this theorem with the same example f(x) = 2x and g(x) = $\frac{x}{2}$, selecting a as 4.

b = f(4) = 8

g'(x) = $\frac{1}{2}$ and hence g'(8) = $\frac{1}{2}$

f'(x) = 2 and hence f'(4) = 2. Now f'(4) = $\frac{1}{2}$ .

Thus, the theorem is verified.

Inverse Sine Function

Let us now take a closer look at the inverse sine function. Let us consider their simplest form as f(x) = sin (x) and g(x) = $\sin^{-1}(x)$.

The sine function f(x) has a domain of all real numbers. But its range is only [-1, 1].
In case of an inverse sine function g(x), the domain gets shrunk to [-1, 1]. On the other hand, the range becomes all real numbers.
There is a divergent opinion on the domain of sine functions and the range of the inverse sine function. Although they are all real numbers, the principal angles correspond to 1 and -1 are only $\frac{p}{2}$ and $\frac{–p}{2}$. Hence according to some sections, this is the domain for sine functions and the range for inverse sine functions.

Inverse Functions Examples


The square function in the positive domain and the principal square root function are inverse to each other. That is,

If f(x) = x2, for x ≥ 0, and if g(x) = $\sqrt{x}$ for x ≥ 0, then f(x) and g(x) are inverse to each other.

The logarithmic function in its domain and the exponential function are inverse to each other. That is,

If f(x) = ln (x), for x ≥ 0, and if g(x) = ex for x ≥ 0, then f(x) and g(x) are inverse to each other.

If f(x) = $\log (x)$, for x ≥ 0, and if g(x) = 10x for x ≥ 0, then f(x) and g(x) are inverse to each other.

Or more explicitly, if f(x) = logb (x), for x ≥ 0, and if g(x) = bx for x ≥ 0, then f(x) and g(x) are inverse to each other.

In trigonometric functions, there is no clear definition on inverse functions. Hence in general they are represented in a general form.

For example, the inverse of sin (x) is represented as sin-1 (x). But both are defined functions.

Derivatives of the Inverse Function

The derivatives of inverse functions are easily worked out by first finding their inverse functions and then differentiating them with respect to the variable. For example, the derivative of the inverse function of f(x) = 2x + 1 is evaluated like this.

If f(x)= 2x + 1, then, f-1(x) = say, g(x) = $\frac{(x – 1)}{2}$, therefore the derivative of f-1(x)= g’(x) = $\frac{1}{2}$.

We will reserve this result for a comment later.

In general, if g(x) = f-1(x), then, f[g(x)] = x

By implicit differentiation, g’(x)*f ‘(x) = 1 or, g’(x) = $[\frac{1}{(f’(x)}]$

We will check back this conclusion with the example we discussed just before that is if f(x) = 2x + 1.
We figured out that f-1(x) = say, g(x) = $\frac{(x – 1)}{2}$

Now as per implicit differentiation,

g’(x) = $[\frac{1}{(f’(x)}]$

= $[\frac{1}{2}]$, because if f(x) = 2x + 1, then f ‘(x) = 2.

Therefore, in cases like this, the simpler method is finding the inverse function and differentiating it.

Now let us take the case of an inverse trigonometric function, say, y = tan-1 (x). or tan y = x

By implicit differentiation, sec2 (y)* y’ = 1 or, [1 + tan2 (y)]*y’ = 1 or, [1 + x2]*y’ = 1 or, y’ = $\frac{1}{(1 + x^{2})}$

Therefore, the derivative of tan-1 (x) = $\frac{1}{(1 + x^{2})}$

Properties of Inverse Functions


There are two important properties of inverse functions.
If f(x) and g(x) are inverse functions, then (fog)(x) = (gof)(x) = x.

If f(x) and g(x) are inverse functions and if both of them are basic functions, then they are symmetric over the axis f(x) = x.
We will show an example graph in the next section.

Graphing Inverse Functions

We already explained that if f(x) and g(x) are inverse to each other, the ordered pairs at any point are interchanged. This concept helps us in graphing inverse functions.
Let, f(x) = 2x and therefore f-1 (x) = $\frac{x}{2}$

Let us consider two values. Since both the given function and its inverse are linear, two points are sufficient to graph both the functions.

If, x = -4, then f(x) = -8 and if, x = 4, then f(x) = 8

Thus, the two ordered pairs of f(x) are (-4, -8) and (4, 8)

By interchanging the ordered pairs, that is, (-8, -4) and (8, 4), we find that they exactly fit in as ordered pairs of f-1 (x). Plot these points and draw the graph as shown below.

Graphing Inverse Functions

It is also noticed from the above graph that the graphs of f(x) and f-1 (x). are symmetrical over the
line f(x) = x.

Topics in Inverse Functions