Lesson 15: Inverse Functions and Logarithms
31
. . . . . . Section 3.2 Inverse Functions and Logarithms V63.0121.034, Calculus I October 21, 2009 Announcements I Midterm course evaluations at the end of class . . Image credit: Roger Smith
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The inverse of a function "undoes" the effect of the function. We look at the implications of that property in the derivative, as well as logarithmic functions, which are inverses of exponential functions.
Transcript of Lesson 15: Inverse Functions and Logarithms
. . . . . .
Section3.2InverseFunctionsandLogarithms
V63.0121.034, CalculusI
October21, 2009
Announcements
I 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
. . . . . .
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
