Question

For the function f, of, x, equals, start fraction, left parenthesis, x, plus, 2, right parenthesis, divided by, 5, end fractionf(x)=
5
(x+2)
​
, find f, to the power minus 1 , left parenthesis, x, right parenthesisf
−1
(x).
Answer
Attempt 1 out of 2
Multiple Choice Answers
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, 5, x, plus, 2f
−1
(x)=5x+2
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, start fraction, x, divided by, 5, end fraction, minus, 2f
−1
(x)=
5
x
​
−2
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, 5, left parenthesis, x, minus, 2, right parenthesisf
−1
(x)=5(x−2)
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, 5, x, minus, 2f
−1
(x)=5x−2

Answer 1

f(x) = (x+2)/5. Let y = (x+2)/5, swap x and y: x = (y+2)/5. Solve for y: y = 5x − 2. So f⁻¹(x) = 5x − 2.