1. Inverse Functions.pdf

7/26/2019 1. Inverse Functions.pdf

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Lecture 1 : Inverse functions

One-to-one Functions A function f is one-to-one if it never takes the same value twice or

f(x1) =f(x2) whenever x1=x2.

Example The function f(x) =x is one to one, because ifx1=x2, then f(x1) =f(x2).On the other hand the function g(x) =x2 is not a one-to-one function, because g(1) =g(1).Graph of a one-to-one function Iff is a one to one function then no two points (x1, y1), (x2, y2)

have the same y-value. Therefore no horizontal line cuts the graph of the equationy = f(x) more thanonce.

Example Compare the graphs of the above functions

Determining if a function is one-to-one

Horizontal Line test: A graph passes the Horizontal line test if each horizontal line cuts the graphat most once.

Using the graph to determine iff is one-to-oneA functionfis one-to-one if and only if the graph y = f(x) passes the Horizontal Line Test.

ExampleWhich of the following functions are one-to-one?

Using the derivative to determine iff is one-to-oneA function whose derivative is always positive or always negative is a one-to-one function. Why?

Example Is the functiong(x) = 4x+ 4 a one-to-one function?

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Inverse functions

Inverse FunctionsIffis a one-to-one function with domain A and rangeB, we can define an inversefunctionf1 (with domain B ) by the rule

f1(y) =x if and only if f(x) =y.

This is a sound definition of a function, precisely because each value of y in the domain of f1 has

exactly onexin A associated to it by the rule y = f(x).

Example Iff(x) =x3 + 1, use the equivalence of equations given above find f1(9) andf1(28).

Notethat the domain off1 equals the range offand the range off1 equals the domain off.

Example Let g(x) =

4x+ 4. What is Domain f?

What is Range g?Doesg1 exist?

What is Domain g1?

What is Range g1?

What is g1(4)?

Finding a Formula For f1(x)

Given a formula for f(x), we would like to find a formula for f1(x). Using the equivalence

x= f1

(y) if and only if y= f(x)

we can sometimes find a formula for f1 using the following method:

1. In the equationy= f(x), if possible solve for x in terms ofy to get a formula x = f1(y).

2. Switch the roles ofx and y to get a formula for f1 of the form y = f1(x).

Example Let f(x) = 2x+1x3

, find a formula for f1(x).

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Composing f and f1.

We haveif x= f1(y) then y= f(x).

Substitutingf(x) fory in the equation on the left, we get

f1(f(x)) = x.

Similarlyif x= f(y) then y= f1(x)

and substituting f1(x) fory in the equation on the left, we get

f(f1(x)) = x.

Example Above, we found that iff(x) = 2x+1x3

, then f1(x) = 3x+1x2

. We can check the above formulafor the composition:

f(f1(x)) =f3x+ 1

x 2

=2

3x+1x2

+ 1

3x+1x2

3

= (6x+ 2 +x 2)/(x 2)(3x+ 1 3x+ 6)/(x 2)=

7x

7 =x.

You should also check thatf1(f(x)) =x.

Graph ofy= f1(x)

Since the equation y =f1(x) is the same as the equation x= f(y), the graphs of both equations areidentical. To graph the equationx= f(y), we note that this equation results from switching the rolesofx and y in the equation y = f(x). This transformation of the equation results in a transformation ofthe graph amounting to reflection in the line y = x. Thus

the graph ofy = f1(x) is a reflection of the graph ofy = f(x) in the line y= x and vice versa.

Note The reflection of the point (x1, y1) n the line y =x is (y1, x1). Therefore if the point (x1, y1) ison the graph ofy = f1(x), we must have (y1, x1) on the graph ofy = f(x).

The graphs off(x) = 2x+1x3

andf1(x) = 3x+1x2

are shown below.

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We can derive properties of the graph ofy = f1(x) from properties of the graph ofy = f(x), sincethey are refections of each other in the line y = x. For example:

TheoremIffis a one-to-one continuous function defined on an interval, then its inverse f1 is alsoone-to-one and continuous. (Thus f1(x) has an inverse, which has to be f(x), by the equivalenceof equations given in the definition of the inverse function.)

Theorem Iffis a one-to-one differentiable function with inverse function f1 andf(f1(a))

= 0,

then the inverse function is differentiable at aand

(f1)(a) = 1

f(f1(a)).

proofy=f1(x) if and only ifx= f(y). Using implicit differentiation we differentiatex= f(y) withrespect to x to get

1 =f(y)dy

dx or

1

f(y)=

dy

dx

or

1

f(y) = (f1

)

(x) or

1

f(f1(x))= (f1

)

(x)

Geometrically this means that if (a, f1(a)) is a point on the curve y = f1(x), then the point(f1(a), a) is on the curvey = f(x) and the slope of the tangent to the curve y = f1(x) at (a, f1(a))is the reciprocal of the tangent to the curve y = f(x) at the point (f1(a), a). The graphs of the functionf(x) = 2x+1

x3 andf1(x) = 3x+1

x2are shown below. You can verify that7 = (f1)(3) = 1

f(10).

NoteTo use the above formula for (f1)(a), you do not need the formula for f1(x), you only needthe value off1 ataand the value off atf1(a).

Example Consider the functionf(x) =

4x+ 4 defined above. Find (f1)(4).

What does the formula from the theorem say?

Use the equivalence of the equations y = f1(x) andx = f(y) to find f1(4).

Put this in the formula from the theorem to find (f1)(4).

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Example Letf(x) =x3 + 1, find (f1)(28).

Example Iffis a one-to-one function with the following properties:

f(10) = 21, f(10) = 2, f1(10) = 4.5, f(4.5) = 3.

Find (f1)(10).

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1. Inverse Functions.pdf