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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.
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. . . . . .
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
. . . . . .
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