On-line Math 21

Inverse Functions

What do we mean by an inverse function? The inverse of a function f (don't write as y = f(x) for now, it adds to the confusion) is another function, g , so that g(f(x)) = x , that is, g undoes what f does. Also, it works the other way for the same functions, f(g(x)) = x . We usually call that g by f^-1 , even though that is not the same as 1/f(x) as you might think. Another way of writing the relationship between f and f^-1 is that

x = f-1(y) Û  y = f(x).

The square root is the standard example of an inverse function.

f-1(x) = Öx

certainly undoes what f(x) = x² does, since

Ö   x2   = x.

Right? Not exactly. If x is positive, then yes, it works that way. This example points out one confusing fact about some inverse functions. Often they don't really make sense. In order for an inverse function to f to exist, not only for each x can there only be one y with y = f(x) (the vertical line test for it to be a function!), but it also has to work the other way around, that is, for each y there can be only one x . This is the horizontal line test.

Now, for the square function f(x) = x² , it fails the horizontal line test. So, no inverse, move on. But, if you restrict the domain of f to be only the nonnegative reals, x ³ 0 , then, for each y = x² , there is only one x , and that x is the inverse applied to y . This described the square root function, the inverse of the squaring function restricted to positive numbers.

x and y

There is a cheap trick the textbooks have at this point. and that is to mess with what is x and what is y. If y = f(x) is the formula for f , then how do you write the formula for f^-1 , the inverse of f , if it exists? One way is to just treat the inverse g as any other function,

y = g(x) = f-1(x),

so that the x is the independent variable, g(x) = y . But x and y play different roles here, for the function f , given x you describe the y , and it's the same for g . But, g ought to be really defined on the range of f (domain of g is the range of f , and the other way around). So maybe it makes more sense to deal with g as g(y) = x . This has the added benefit of making sense when you solve. However, we are very used to thinking of x as the independent variable, so, usually, we will write the inverse as a function of x. As the next paragraph shows, often that means simply switching the y's and x's as a last step when finding the inverse.

Finding the inverse

Given y = f(x) , how do you find the inverse g , if it does exist (after you straighten out the domain of f so it works)? Well, the easy way is to find it as x = f^-1(y) . That way, x plays the same role throughout. Thinking along those lines, if y = f(x) , then for that x and that y , f^-1(y) = x , since f^-1 undoes what f does, so you go forward and back, x® y, y® x . The method is to simply solve the equation y = f(x) for y in terms of x. If the solution makes sense, the inverse exists, and there is the formula.

Example 1 If

f(x) = 2x+1 x-1 ,

find the formula for the inverse function g = f^-1 .

Solution

Exercise 1 Find the inverse of f(x) = x³-1 .

f-1(x) =

Derivatives of inverse functions; implicit differentiation

The basic technique to find derivatives of an inverse function f^-1(x) in terms of the derivative of the original function f(x) is one case of a more general trick, called implicit differentiation.

Some functions are not given explicitly as a formula, but implicitly in that the function (usually called y ) appears in an equation involving it and the variable x . The prototype is the equation

x2+y2 = 1.

y is defined by this as the function which takes a given x to the number y that fits with x in that equation. Of course, in this case we can solve for y explicitly,

y = ± Ö   1-x2   .

±? That's not a function. But, in a small enough region, (from x = -1 to x = 1 ), starting at (0,1) , say,

y = Ö   1-x2   .

On the other hand, in a small enough region near (0,-1) ,

y = - Ö   1-x2   .

This is a standard situation with functions defined implicitly: an equation in x and y will define a function y = f(x) for x in a neighborhood so the curve passes the vertical line test for x 's in that neighborhood with some value of y starting with a given point (x₀,y) .

It's really quite easy to differentiate these functions; in fact it's often easier to differentiate them than it is to solve for them explicitly. Just differentiate both sides of the equation, using the chain rule.

x2+y2 = 1  Think of this as: x2+(y(x))2 = 1.  Now, differentiate both sides: 2x+2(y(x)) dy dx = 0 2x+2y dy dx = 0 2y dy dx = -2x dy dx = -2x/2y = -x/y.

Exercise 2 Find

dy dx

at (2,1) using implicit differentiation, if: y³x+yx²+y⁵ = 11 .

dy dx ê ê ê (2,1)  =

In the particular case of an inverse

y = f-1(x)

to a function f(x) , which we assume we already know how to differentiate, we can write the equation

y = f-1(x)

as

f(y) = x

which is an equation defining y implicitly as a function of x , and it defined y as f^-1(x) , of course.

Now differentiate both sides implicitly,

f(y) = x f¢(y) dy dx = 1,

and solve for dy/dx -which is the derivative of the inverse:

( f-1(x)) ¢ = dy dx = 1 f¢(y) ,

where that y in the last denominator can be re-written as f^-1(x) ,

( f-1(x)) ¢ = 1 f¢( f-1(x)) .

In the applications of this to particular functions, later in this chapter, we will often be able to simplify that last expression considerably.

Example 2 Use implicit differentiation to find the derivative of

Öx.

Solution

Exercise 3 If f(x) is a function whose derivative satisfies the equation

f¢(x) = ( f(x)) 2,

find the derivative of f^-1(x) .

( f-1(x)) ¢ =

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