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Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 6.2
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Exercise 136 Page 295

Practice makes perfect
a

We will begin by finding the inverse of p(x). First, we need to replace p(x) with y. From there, we switch x and y and solve for y.

y=3( x^3+6) → x=3( y^3+6)The resulting equation will be the inverse of the given function.

x=3(y^3+6)
â–¼
Solve for y
DivEqn

.LHS /3.=.RHS /3.

x/3=y^3+6
SubEqn

LHS-6=RHS-6

x/3-6=y^3
RadicalEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x/3-6)=sqrt(y^3)
RootPowToNumber

sqrt(a^n)=a

sqrt(x/3-6)=y
RearrangeEqn

Rearrange equation

y=sqrt(x/3-6)

Finally, to indicate that this is the inverse function of p(x), we will replace y with p^(- 1)(x). p^(- 1)(x)=sqrt(x/3-6)

b

We will begin by finding the inverse of k(x). First, we need to replace k(x) with y. From there, we switch x and y and solve for y.

y=3 x^3+6 → x=3 y^3+6The resulting equation will be the inverse of the given function.

x=3y^3+6
â–¼
Solve for y
SubEqn

LHS-6=RHS-6

x-6=3y^3
DivEqn

.LHS /3.=.RHS /3.

x-6/3=y^3
RadicalEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x-6/3)=sqrt(y^3)
RootPowToNumber

sqrt(a^n)=a

sqrt(x-6/3)=y
RearrangeEqn

Rearrange equation

y=sqrt(x-6/3)

Finally, to indicate that this is the inverse function of k(x), we will replace y with k^(- 1)(x). k^(- 1)(x)=sqrt(x-6/3)

c

We will begin by finding the inverse of h(x). First, we need to replace h(x) with y. From there, we switch x and y and solve for y.

y=x+1/x-1 → x=y+1/y-1The resulting equation will be the inverse of the given function.

x=y+1/y-1
â–¼
Solve for y
MultEqn

LHS * (y-1)=RHS* (y-1)

x(y-1)=y+1
Distr

Distribute x

xy-x=y+1
AddEqn

LHS+x=RHS+x

xy=y+x+1
SubEqn

LHS-y=RHS-y

xy-y=x+1
FactorOut

Factor out y

y(x-1)=x+1
DivEqn

.LHS /(x-1).=.RHS /(x-1).

y=x+1/x-1

Finally, to indicate that this is the inverse function of h(x), we will replace y with h^(- 1)(x). h^(- 1)(x)=x+1/x-1

d

We will begin by finding the inverse of j(x). First, we need to replace j(x) with y. From there, we switch x and y and solve for y.

y=2/3- x → x=2/3- yThe resulting equation will be the inverse of the given function.

x=2/3-y
â–¼
Solve for y
MultEqn

LHS * (3-y)=RHS* (3-y)

x(3-y)=2
DivEqn

.LHS /x.=.RHS /x.

(3-y)=2/x
RemovePar

Remove parentheses

3-y=2/x
SubEqn

LHS-3=RHS-3

- y=-3+2/x
MultEqn

LHS * (-1)=RHS* (-1)

y=3-2/x

Finally, to indicate that this is the inverse function of f(x), we will replace y with j^(- 1)(x). j^(- 1)(x)=3-2/x

Subchapter links expand_more
arrow_right 6.2.1 Using Logarithms to Solve Exponential Equations p.284-285
95
96
97
98
Using Logarithms to Solve Exponential Equations 99 (Page 284)
100
Using Logarithms to Solve Exponential Equations 101 (Page 285)
Using Logarithms to Solve Exponential Equations 102 (Page 285)
103
arrow_right 6.2.2 Investigating the Properties of Logarithms p.290-291
113
114
115
116
117
118
Investigating the Properties of Logarithms 119 (Page 291)
Investigating the Properties of Logarithms 120 (Page 291)
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122
arrow_right 6.2.3 Writing Equations of Exponential Functions p.294-295
127
128
129
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131
Writing Equations of Exponential Functions 132 (Page 295)
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134
Writing Equations of Exponential Functions 135 (Page 295)
Writing Equations of Exponential Functions 136 (Page 295)
arrow_right 6.2.4 An Application of Logarithms p.299-301
An Application of Logarithms 138 (Page 299)
An Application of Logarithms 139 (Page 299)
An Application of Logarithms 140 (Page 299)
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An Application of Logarithms 145 (Page 300)
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An Application of Logarithms 147 (Page 301)