Let $f(x) = ax + b$ for $0 \le x \le 1$, where $a$ and $b$ are constants such that $a < 0$. What is the inverse of $f(x)$ in terms of $a$ and $b$?
3 answers
We see that $f(0) = b$. Also, $f(1) = a + b$. Solving for $a$, we find $a = f(1) - f(0) = (a + b) - b = a$, so $f(1) = 0$. Hence, the inverse of $f(x)$ is given by \[f^{-1}(x) = \boxed{-ax + b}.\]
it doesn't really matter whether a and b are negative.
If f(x) = ax+b then
f-1(x) = (x-b)/a
or, f-1(x) = 1/a x - b/a
it's easy to check, since we know that
f-1(f(x)) = f(f-1(x)) = x
f(f-1(x)) = a[(x-b)/a)+b = x
f-1(f(x)) = (f(x)-b)/a+b = (ax+b-b)/a = x
If f(x) = ax+b then
f-1(x) = (x-b)/a
or, f-1(x) = 1/a x - b/a
it's easy to check, since we know that
f-1(f(x)) = f(f-1(x)) = x
f(f-1(x)) = a[(x-b)/a)+b = x
f-1(f(x)) = (f(x)-b)/a+b = (ax+b-b)/a = x
Oops! You are correct. I apologize for the mistake in my previous response. The correct inverse function is indeed $f^{-1}(x) = \frac{x - b}{a}$. Thank you for pointing that out.