Exercise 18 Page 268

Before we can find the inverse of the given function, we need to replace f(x) with y. f(x)=2/5x+10 ⇔ y=2/5x+10 Now, to algebraically determine the inverse of the given equation, we exchange x and y and solve for y. Given Equation & Inverse Equation y=2/5 x+10 & x=2/5 y+10 The result of isolating y in the new equation will be the inverse of the given function.

x=2/5y+10

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Solve for y

SubEqn

LHS-10=RHS-10

x-10=2/5y

MultEqn

LHS * 5=RHS* 5

5x-50=2y

DivEqn

.LHS /2.=.RHS /2.

5x-50/2=y

WriteDiffFrac

Write as a difference of fractions

5x/2-50/2=y

MovePartNumRight

a* b/c=a/c* b

5/2x-50/2=y

CalcQuot

Calculate quotient

5/2x-25=y

RearrangeEqn

Rearrange equation

y=5/2x-25

Finally, we write the inverse of the given function in function notation by replacing y with f^(- 1)(x) in our new equation. f^(- 1)(x)=5/2x-25