. . . . . .

Section3.2InverseFunctionsandLogarithms

V63.0121.027, CalculusI

October22, 2009

Announcements

I Quizon§§2.5–2.6nextweekI Midtermcourseevaluationsattheendofclass

..Imagecredit: RogerSmith

. . . . . .

Outline

InverseFunctions

DerivativesofInverseFunctions

LogarithmicFunctions

. . . . . .

Whatisaninversefunction?

DefinitionLet f beafunctionwithdomain D andrange E. The inverse of f isthefunction f−1 definedby:

f−1(b) = a,

where a ischosensothat f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

. . . . . .

Whatisaninversefunction?

DefinitionLet f beafunctionwithdomain D andrange E. The inverse of f isthefunction f−1 definedby:

f−1(b) = a,

where a ischosensothat f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

. . . . . .

Whatfunctionsareinvertible?

Inorderfor f−1 tobeafunction, theremustbeonlyone a in Dcorrespondingtoeach b in E.

I Suchafunctioniscalled one-to-oneI Thegraphofsuchafunctionpassesthe horizontallinetest:anyhorizontallineintersectsthegraphinexactlyonepointifatall.

I If f iscontinuous, then f−1 iscontinuous.

. . . . . .

Graphinganinversefunction

I Thegraphof f−1

interchangesthe x and ycoordinateofeverypointonthegraphof f

I Theresultisthattogetthegraphof f−1, weneedonlyreflectthegraphof f inthediagonalline y = x.

.

.f

.f−1

. . . . . .

Graphinganinversefunction

I Thegraphof f−1

interchangesthe x and ycoordinateofeverypointonthegraphof f

I Theresultisthattogetthegraphof f−1, weneedonlyreflectthegraphof f inthediagonalline y = x.

.

.f

.f−1

. . . . . .

Howtofindtheinversefunction

1. Write y = f(x)

2. Solvefor x intermsof y

3. Toexpress f−1 asafunctionof x, interchange x and y

ExampleFindtheinversefunctionof f(x) = x3 + 1.

Answery = x3 + 1 =⇒ x = 3

√y− 1, so

f−1(x) = 3√x− 1

. . . . . .

Howtofindtheinversefunction

1. Write y = f(x)

2. Solvefor x intermsof y

3. Toexpress f−1 asafunctionof x, interchange x and y

ExampleFindtheinversefunctionof f(x) = x3 + 1.

Answery = x3 + 1 =⇒ x = 3

√y− 1, so

f−1(x) = 3√x− 1

. . . . . .

Howtofindtheinversefunction

1. Write y = f(x)

2. Solvefor x intermsof y

3. Toexpress f−1 asafunctionof x, interchange x and y

ExampleFindtheinversefunctionof f(x) = x3 + 1.

Answery = x3 + 1 =⇒ x = 3

√y− 1, so

f−1(x) = 3√x− 1

. . . . . .

Outline

InverseFunctions

DerivativesofInverseFunctions

LogarithmicFunctions

. . . . . .

derivativeofsquareroot

Recallthatif y =√x, wecanfind

dydx

byimplicitdifferentiation:

y =√x =⇒ y2 = x

=⇒ 2ydydx

= 1

=⇒ dydx

=12y

=1

2√x

Notice 2y =ddy

y2, and y istheinverseofthesquaringfunction.

. . . . . .

Theorem(TheInverseFunctionTheorem)Let f bedifferentiableat a, and f′(a) ̸= 0. Then f−1 isdefinedinanopenintervalcontaining b = f(a), and

(f−1)′(b) =1

f′(f−1(b))

“Proof”.If y = f−1(x), then

f(y) = x,

Sobyimplicitdifferentiation

f′(y)dydx

= 1 =⇒ dydx

=1

f′(y)=

1

f′(f−1(x))

. . . . . .

Theorem(TheInverseFunctionTheorem)Let f bedifferentiableat a, and f′(a) ̸= 0. Then f−1 isdefinedinanopenintervalcontaining b = f(a), and

(f−1)′(b) =1

f′(f−1(b))

“Proof”.If y = f−1(x), then

f(y) = x,

Sobyimplicitdifferentiation

f′(y)dydx

= 1 =⇒ dydx

=1

f′(y)=

1

f′(f−1(x))

. . . . . .

Outline

InverseFunctions

DerivativesofInverseFunctions

LogarithmicFunctions

. . . . . .

Logarithms

Definition

I Thebase a logarithm loga x istheinverseofthefunction ax

y = loga x ⇐⇒ x = ay

I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.

Facts

(i) loga(x · x′) = loga x + loga x

′

(ii) loga( xx′

)= loga x− loga x

′

(iii) loga(xr) = r loga x

. . . . . .

Logarithms

Definition

I Thebase a logarithm loga x istheinverseofthefunction ax

y = loga x ⇐⇒ x = ay

I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.

Facts

(i) loga(x · x′) = loga x + loga x

′

(ii) loga( xx′

)= loga x− loga x

′

(iii) loga(xr) = r loga x

. . . . . .

Logarithms

Definition

I Thebase a logarithm loga x istheinverseofthefunction ax

y = loga x ⇐⇒ x = ay

I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.

Facts

(i) loga(x · x′) = loga x + loga x

′

(ii) loga( xx′

)= loga x− loga x

′

(iii) loga(xr) = r loga x

. . . . . .

Logarithms

Definition

I Thebase a logarithm loga x istheinverseofthefunction ax

y = loga x ⇐⇒ x = ay

I Thenaturallogarithm ln x istheinverseof ex. Soy = ln x ⇐⇒ x = ey.

Facts

(i) loga(x · x′) = loga x + loga x

′

(ii) loga( xx′

)= loga x− loga x

′

(iii) loga(xr) = r loga x

. . . . . .

Logarithmsconvertproductstosums

I Suppose y = loga x and y′ = loga x′

I Then x = ay and x′ = ay′

I So xx′ = ayay′ = ay+y′

I Therefore

loga(xx′) = y + y′ = loga x + loga x

′

. . . . . .

ExampleWriteasasinglelogarithm: 2 ln 4− ln 3.

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Writeasasinglelogarithm: ln34

+ 4 ln 2

Answerln 12

. . . . . .

ExampleWriteasasinglelogarithm: 2 ln 4− ln 3.

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Writeasasinglelogarithm: ln34

+ 4 ln 2

Answerln 12

. . . . . .

ExampleWriteasasinglelogarithm: 2 ln 4− ln 3.

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Writeasasinglelogarithm: ln34

+ 4 ln 2

Answerln 12

. . . . . .

ExampleWriteasasinglelogarithm: 2 ln 4− ln 3.

Solution

I 2 ln 4− ln 3 = ln 42 − ln 3 = ln42

3

I notln 42

ln 3!

Example

Writeasasinglelogarithm: ln34

+ 4 ln 2

Answerln 12

. . . . . .

..“lawn”

.

.Imagecredit: Selva

. . . . . .

Graphsoflogarithmicfunctions

. .x

.y.y = 2x

.y = log2 x

. .(0, 1)

..(1, 0)

.y = 3x

.y = log3 x

.y = 10x

.y = log10 x

.y = ex

.y = ln x

. . . . . .

Graphsoflogarithmicfunctions

. .x

.y.y = 2x

.y = log2 x

. .(0, 1)

..(1, 0)

.y = 3x

.y = log3 x

.y = 10x

.y = log10 x

.y = ex

.y = ln x

. . . . . .

Graphsoflogarithmicfunctions

. .x

.y.y = 2x

.y = log2 x

. .(0, 1)

..(1, 0)

.y = 3x

.y = log3 x

.y = 10x

.y = log10 x

.y = ex

.y = ln x

. . . . . .

Graphsoflogarithmicfunctions

. .x

.y.y = 2x

.y = log2 x

. .(0, 1)

..(1, 0)

.y = 3x

.y = log3 x

.y = 10x

.y = log10 x

.y = ex

.y = ln x

. . . . . .

Changeofbaseformulaforexponentials

FactIf a > 0 and a ̸= 1, then

loga x =ln xln a

Proof.

I If y = loga x, then x = ay

I So ln x = ln(ay) = y ln aI Therefore

y = loga x =ln xln a

. . . . . .

Changeofbaseformulaforexponentials

FactIf a > 0 and a ̸= 1, then

loga x =ln xln a

Proof.

I If y = loga x, then x = ay

I So ln x = ln(ay) = y ln aI Therefore

y = loga x =ln xln a