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If the function f:R->R satisfies f(2x-1)=5x+2, find f^(-1) (3x+1)?
given answer: (6x-7) /5
3 Answers
- mizooLv 72 years agoFavorite Answer
2x - 1 = a => x = (1/2)(a + 1)
f(a) = 5x + 2
f(a) = 5((1/2)(a + 1)) + 2
f(a) = (5/2)a + 9/2
f(x) = (5/2)x + 9/2
f^(-1)(x) :
y = (5/2)x + 9/2
y - 9/2 = 5/2 * x
x = 2/5 * (y - 9/2)
f^-1(x) = 2/5 * (x - 9/2)
f^-1(3x + 1) = 2/5 * ((3x + 1) - 9/2)
f^-1(3x + 1) = 6/5 x + 2/5 - 9/5
f^-1(3x + 1) = (6x - 7)/5
- ?Lv 72 years ago
We first get f(x)
Since f(2x-1) = 5x+2
set u = 2x-1 or x = (u+1)/2
and we have f(u) = 5(u+1)/2 + 2 = (5u+ 9)/2
or y = (5u+9)/2
to find the inverse we switch y and u
u = (5y+9)/2
y = ( 2u-9)/5
or f^-1(u) = (2u-9/5
So f^-1(3x+1) = ( 2 (3x+1) - 9)/5 = ( 6x-7)/5
- RealProLv 72 years ago
Well since (5/2)(2x-1) + 9/2 = 5x+2, then f(x) = (5/2)x + 9/2
and f^-1(x) = (2x-9)/5
So f^-1(3x+1) = (2(3x+1)-9)/5 = (6x-7)/5
