A-Level Functions: worked solution

Question

The function f is f: x → 2x − 1/(2x) for 0<x<2, and f⁻¹ is known to exist. Express f⁻¹ in a similar form.

f: x -> 2x - 1/(2x), for 0 < x < 2. Step 1: Set y equal to the rule and make x the subject. y = 2x - 1/(2x) The unknown x appears both on its own and in a denominator, so clear the fraction by multiplying every term by 2x (allowed, since 0 < x < 2 guarantees x is not 0): 2xy = 4x² - 1 Bring everything to one side so that it is a QUADRATIC in x: 4x² - 2xy - 1 = 0 Step 2: Solve that quadratic for x, treating y as a constant. Use the formula with a = 4, b = -2y, c = -1: x = [2y +/- sqrt((-2y)² - 4(4)(-1))]/(2 x 4) = [2y +/- sqrt(4y² + 16)]/8 Take the factor 4 out of the square root: sqrt(4y² + 16) = sqrt(4(y² + 4)) = 2 sqrt(y² + 4), so = [2y +/- 2 sqrt(y² + 4)]/8 = [y +/- sqrt(y² + 4)]/4 Step 3: Decide which sign to keep. An inverse function must give ONE value. The domain of f is 0 < x < 2, so x must be positive. Since y² + 4 > y², we have sqrt(y² + 4) > |y|, so y + sqrt(y² + 4) > 0 and y - sqrt(y² + 4) < 0. The minus sign would make x negative, which is outside the domain, so reject it: x = [y + sqrt(y² + 4)]/4 = y/4 + (1/4)sqrt(y² + 4). Step 4: Find the domain of the inverse, which is the RANGE of f. f'(x) = 2 + 1/(2x²), which is positive for every x in (0, 2), so f is strictly increasing there. As x -> 0+, the term 1/(2x) -> +infinity, so f(x) -> -infinity. At the excluded endpoint x = 2, f(2) = 2(2) - 1/(2 x 2) = 4 - 1/4 = 15/4. Since f is increasing and 2 is not included, the range is R_f = (-infinity, 15/4), and this is the domain of f⁻¹. Step 5: Write the answer in the same form as f, renaming the variable x. f⁻¹: x -> x/4 + (1/4)sqrt(x² + 4), x < 15/4.