The inverse of f (x), denoted f −1(x), is the function that reverses the effect of f (x).

For example, the inverse of f (x) = x3 is the cube root function f −1(x) = x1/3. Given a table of function values for f (x), we obtain a table for f −1(x) by interchanging the x and y columns:

DEFINITION Inverse Let f (x) have domain D and range R. If there is a function g (x) with domain R such that

then f (x) is said to be invertible. The function g (x) is called the inverse function and is denoted f −1(x).

Show that f (x) = 2x − 18 is invertible. What are the domain and range of f −1(x)?

1

2

2

18

18

xf

x y

xy

y x

1 is linear the & is D Rf x

So when does a function f (x) have an inverse? The answer is: When f (x) is one-to-one, which means that f (x) takes on each value in its range at most once.

Which graphs are not one-to-one?

a b

, but f a f b b a

a b

, but f a f b b a

When f (x) is one-to-one on its domain D, the inverse function f −1(x) exists and its domain is equal to the range R of f. Indeed, for every c R, there is precisely one element a D such that f (a) = c and we may define f −1(c) = a. With this definition, f (f −1(c)) = f (a) = c and f −1(f (a)) = f −1(c) = a. This proves the following theorem.

THEOREM 1 Existence of Inverses The inverse function f −1(x) exists if and only if f (x) is one-to-one on its domain D. Furthermore,

• Domain of f = range of f −1.• Range of f = domain of f −1.

1

3 2

5 1

5 1 3 2

5 3 2

5 3 2

2

5 3

yx

y

x y y

xy x y

xy y

xf x

x

yx

Show that is invertible. Find f −1 & determine the domain and range of f and f −1.

1 3: :{ / } :{ / }

5 5f D x x R y y

1 3 1: :{ / } :{ / }

5 5f R x x R y y

is one-to-one is invertiblef f Horizontal Line Test

Often, it is impossible to find a formula for the inverse because we cannot solve for x explicitly in the equation y = f (x). For example, the function f(x) = x + ex has an inverse, but we must make do without an explicit formula for it.

Show that f (x) = x5 + 4x + 3 is one-to-one.

• If n odd and c > 0, then cxn is increasing.• A sum of increasing functions is increasing.

' 0 in .f x x

The increasing function f (x) = x5 + 4x + 3 satisfies the Horizontal Line Test.

Restricting the Domain Find a domain on which f (x) = x2 is one-to-one and determine its inverse on this domain.

1f x x

f (x) = x2 satisfies the Horizontal Line Test on the domain {x : x ≥ 0}.

Next we describe the graph of the inverse function. The reflection of a point (a, b) through the line y = x is, by definition, the point (b, a). Note that if the x-and y-axes are drawn to the same scale, then (a, b) and (b, a) are equidistant from the line y = x and the segment joining them is perpendicular to y = x.

The reflection (a, b) through the line y = x is the point (b, a).

The graph of f −1(x) is the reflection of the graph of f (x) through the line y = x.

Sketching the Graph of the Inverse Sketch the graph of the inverse of

4 .f x x y

x

y xy x 4y x

422

2

4y x

1 24g x f x x

THEOREM 2 Derivative of the Inverse Assume that f (x) is differentiable and one-to-one with inverse g(x) = f −1(x). If b belongs to the domain of g(x) and f (g (b)) 0, then g (b) exists and

1

''

g bf g b

''

GRAPHICAL INSIGHT The formula for the derivative of the inverse function has a clear graphical interpretation. Consider a line L of slope m and let L be its reflection through y = x. Then the slope of L is 1/m. Indeed, if (a, b) and (c, d) are any two points on L, then (b, a) and (d, c) lie on L and

''

'

Now recall that the graph of the inverse g (x) is obtained by reflecting the graph of f (x) through the line y = x. As we can see, the tangent line to y = g (x) at x = b is the reflection of the tangent line to y = f (x) at x = a [where b = f (a) and a = g (b)]. These tangent lines have reciprocal slopes, and thus g (b) = 1/f (a) = 1/f (g (b)), as claimed in Theorem 2. '' '

Calculate g (x), where g(x) is the inverse of the function f (x) = x4 + 10 on the domain {x : x ≥ 0}.

'

4 410 10x y y g x x

33' 4 ' 4 'f x x f g x g x g x

3/ 4

1'

'

1

4 10g x

f g x x

31/ 44 10x

We obtain this same result by differentiating g (x) = (x − 10)1/4 directly.

Calculate ' without solving for g x g x

Calculate g (1), where g (x) is the inverse of f (x) = x + ex.

In this case, we cannot solve for g (x) explicitly, but a formula for g (x) is not needed. All we need is the particular value g (1), which we can find by solving f (x) = 1. By inspection, x + ex = 1 has solution x = 0. Therefore, f (0) = 1 and, by definition of the inverse, g (1) = 0. Since f (x) = 1 + ex,

'

1

''

g bf g b

'