f(x) = (5e^x-6)/(17e^x+13) find f^-1(x). since the domain of f^-1(x) is in the open interval (a, b) what is a and and b?
i figured out the f^-1(x) is the ln ((-13x-6)/(17x-5)) i just need a and b plz help before 11 pm
For logarithmic expression, the value of u in ln u should be more than 0. So:
(13x + 6)/(5 - 17x) > 0
Use the number line to solve this, you'll get the result.
If my calculations are correct, then: (-6/13,5/17)
To find the inverse function, f^(-1)(x), of the given function f(x), we need to swap the roles of x and y and solve for y.
1. Begin with the equation: y = (5e^x - 6) / (17e^x + 13).
2. Swap x and y: x = (5e^y - 6) / (17e^y + 13).
3. Multiply both sides of the equation by (17e^y + 13): x(17e^y + 13) = 5e^y - 6.
4. Expand and rearrange the equation: 17xe^y + 13x = 5e^y - 6.
5. Move all terms involving e^y to one side and x to the other side: 17xe^y - 5e^y = -13x - 6.
6. Factor out e^y from the left side: e^y(17x - 5) = -13x - 6.
7. Divide both sides by (17x - 5): e^y = (-13x - 6) / (17x - 5).
8. Take the natural logarithm of both sides to eliminate the exponential term: ln(e^y) = ln((-13x - 6) / (17x - 5)).
9. Simplify the left side: y = ln((-13x - 6) / (17x - 5)).
10. Finally, we have found the inverse function: f^(-1)(x) = ln((-13x - 6) / (17x - 5)).
Regarding the domain of f^(-1)(x), which is denoted as (a, b), we need to consider the domain of the original function f(x) = (5e^x - 6) / (17e^x + 13) to determine the range of f^(-1)(x).
The original function f(x) contains exponential terms and does not have any specific restrictions on its domain. Therefore, the domain of f(x) is the entire real number line, (-∞, ∞).
Since the inverse function is a reflection of the original function, the range of f^(-1)(x) corresponds to the domain of f(x). Thus, the domain of f^(-1)(x) is also the entire real number line, (-∞, ∞), which implies a = -∞ and b = ∞.