A-Level Functions: worked solution

Question

The function f is f(x) = 1 + 3/(e^x - 2) for real x with x > ln 2. Find f inverse (x) and state its domain.

Method: to invert a function, write y = f(x), rearrange to make x the subject, then swap the letters. The domain of the inverse is the RANGE of f, so that has to be worked out too. Step 1 - set y equal to f(x). y = 1 + 3/(e^x - 2) Step 2 - isolate the fraction by subtracting 1 from both sides. y - 1 = 3/(e^x - 2) Step 3 - clear the fraction. Multiply both sides by (e^x - 2), which is not zero because x > ln 2 forces e^x > 2. (y - 1)(e^x - 2) = 3 Now divide both sides by (y - 1) (this is safe, since y > 1 as Step 6 shows). e^x - 2 = 3/(y - 1) Step 4 - make x the subject. Add 2 to both sides, then take natural logarithms. e^x = 2 + 3/(y - 1) x = ln(2 + 3/(y - 1)) Step 5 - swap y for x to write the inverse as a function of x. f inverse (x) = ln(2 + 3/(x - 1)) Step 6 - find the range of f, because that is the domain of f inverse. x > ln 2 gives e^x > 2, so e^x - 2 > 0. As x runs from just above ln 2 up to infinity, e^x - 2 runs through every value in (0, infinity). So 3/(e^x - 2) is large and positive when e^x - 2 is near 0, and shrinks towards 0 as x grows: it takes every value in (0, infinity). Adding the 1 in front, f(x) takes every value greater than 1, so the range of f is y > 1. Step 7 - state the domain of the inverse. Domain of f inverse is x > 1.