SOLUTION: the equation defines a one-to-one function f. f(x) = 4x − 1 Verify that f ∘ f −1 and f −1 ∘ f are both the identity function. (f ∘ f −1)(x) = f(fâ€

Algebra ->  Functions -> SOLUTION: the equation defines a one-to-one function f. f(x) = 4x − 1 Verify that f ∘ f −1 and f −1 ∘ f are both the identity function. (f ∘ f −1)(x) = f(f†     Log On


   



Question 1206605: the equation defines a one-to-one function f.
f(x) = 4x − 1
Verify that
f ∘ f −1
and
f −1 ∘ f
are both the identity function.
(f ∘ f −1)(x) = f(f −1(x))
(f −1 ∘ f)(x) = f −1(f(x))

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29+=+4x+%E2%88%92+1
Verify that
f ∘ f%E2%80%89%5E-1
and
f%5E-1 ∘ f
are both the identity function.

(f+∘ f%5E-1)%28x%29+=+f%28f%5E-1%28x%29%29
(f%5E-1 ∘ f)%28x%29+=+f%5E-1%28f%28x%29%29

find f%5E-1
f%28x%29+=+4x+-+1...f%28x%29+=+y
y=+4x+-1...swap variables
x=+4y+-1...solve for y
x%2B1=+4y
y=x%2F4%2B1%2F4
=>
f%5E-1%28x%29=x%2F4%2B1%2F4

(f+∘ f%5E-1)%28x%29+=f%28f%5E-1%29=f%28x%2F4%2B1%2F4%29=4%28x%2F4%2B1%2F4%29+-+1=x%2B1+-+1=x
(f%5E-1 ∘ f)%28x%29+=f%5E-1%284x-1%29=%284x-1%29%2F4%2B1%2F4=4x%2F4-1%2F4%2B1%2F4=x

proven that
(f+∘ f%5E-1)%28x%29+=+f%28f%5E-1%28x%29%29=x
(f%5E-1 ∘ f)%28x%29+=+f%5E-1%28f%28x%29%29=x