3.1 Question 14

[Eigenvalue]

Precalculus
Michael Sullivan
Edition 4
Chapter 3, Section 3.1

g(x)=5x-4

a) Determine the slope and y-intercept of each function

b) Use the smile and y-intercept to graph the linear function

c) Determine the domain and range of the function

d) Determine the average rate of change of the function

e) Determine whether the linear function is increasing, decreasing, or constant

[Eigenvalue]

f) Find the inverse of g(x)
Let y=g(x) and switch the x and y (the domain of a function is the range of its inverse, and vice versa)

x=5y-4
Add 4 to both sides: x+4=5y
Divide both sides by 5: y=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]
[tex]g^{-1}[/tex]=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]

Check:
For an inverse function, ([tex]g^{-1 }[/tex] o g)(x)=(g o [tex]g^{-1}[/tex])(x)=x

([tex]g^{-1 }[/tex] o g)(x)=g[tex]x^{-1 }[/tex](g(x)) substitute g(x) for the input of [tex]g^{-1 }[/tex](x): [tex]\frac{1}{5}[/tex](5x-4)+[tex]\frac{4}{5}[/tex]=x

(g o [tex]g^{-1 }[/tex])(x)=g([tex]g^{-1 }[/tex](x))
substitute [tex]g^{-1}[/tex] for the input of g
5([tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex])-4=x

[nycmath]

f) Find the inverse of g(x)
Let y=g(x) and switch the x and y (the domain of a function is the range of its inverse, and vice versa)

x=5y-4
Add 4 to both sides: x+4=5y
Divide both sides by 5: y=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]
[tex]g^{-1}[/tex]=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]

Check:
For an inverse function, ([tex]g^{-1 }[/tex] o g)(x)=(g o [tex]g^{-1}[/tex])(x)=x

([tex]g^{-1 }[/tex] o g)(x)=g[tex]x^{-1 }[/tex](g(x)) substitute g(x) for the input of [tex]g^{-1 }[/tex](x): [tex]\frac{1}{5}[/tex](5x-4)+[tex]\frac{4}{5}[/tex]=x

(g o [tex]g^{-1 }[/tex])(x)=g([tex]g^{-1 }[/tex](x))
substitute [tex]g^{-1}[/tex] for the input of g
5([tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex])-4=x

I don't recall inverse functions in Sullivan's 3.1.

[nycmath]

f) Find the inverse of g(x)
Let y=g(x) and switch the x and y (the domain of a function is the range of its inverse, and vice versa)

x=5y-4
Add 4 to both sides: x+4=5y
Divide both sides by 5: y=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]
[tex]g^{-1}[/tex]=[tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex]

Check:
For an inverse function, ([tex]g^{-1 }[/tex] o g)(x)=(g o [tex]g^{-1}[/tex])(x)=x

([tex]g^{-1 }[/tex] o g)(x)=g[tex]x^{-1 }[/tex](g(x)) substitute g(x) for the input of [tex]g^{-1 }[/tex](x): [tex]\frac{1}{5}[/tex](5x-4)+[tex]\frac{4}{5}[/tex]=x

(g o [tex]g^{-1 }[/tex])(x)=g([tex]g^{-1 }[/tex](x))
substitute [tex]g^{-1}[/tex] for the input of g
5([tex]\frac{1}{5}[/tex]x+[tex]\frac{4}{5}[/tex])-4=x

Question 14 has nothing to do with finding inverse functions.

[nycmath]

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[nycmath]

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