Let g(x) = f^-1(x), the inverse of function f. Write the rule for for g(x) for each function f.

Question

a. f(x) = (1/4)^x

b. f(x) = 5x - 2

Answer 1

f(x) = (1/4)^x = y

switch x and y, then solve for the new y

(1/4)^y = x

(think of a simpler example, such as 2^3 = 8, and 2 = cube root of 8. replace 8 with x, 2 with 1/4, 3 with y)

(1/4)^y = x

is equivalent to:

y = f^-1(x) = lnx/ln4 = about .72lnx

y = log base 1/4 of x

convert to log base e

= lnx/ln(1/4)

= -.72ln(.25)

b f(x) =5x -2

switch x and y

x =5y-2

5y= x+2

y = f^-1(x) = (x+2)/5

f(x) multiplies by 5 then subtracts 2

f^-1(x) divides by 5 then adds 2, just the reverse of f(x)

Answer 2

It's a property (definition?) of inverse functions that f( f^-1(x) )=x. So, f(g(x))=x for this problem. This gives us an avenue of finding g(x): replace x with g(x) in the expression for f(x), and then solve for g(x).

For instance, for part a., f(g(x))=(1/4)^g(x)=x. We can use algebra to solve this for g(x)!

(1/4)^g(x)=x

Ln((1/4)^g(x))=Ln(x)

g(x) Ln(1/4)=Ln(x)

g(x)= - Ln(x) / Ln(4)

And for part b.

f(g(x))=5g(x)-2=x

5g(x)=x+2

g(x)=(1/5)(x+2)