A-Level Functions: worked solution

Question

With f: x -> (1/2)sqrt(36 - (x - 3)^2) on k <= x < 9 and k = 3 from part (a), find f^{-1}(x).

With k = 3 the function is f(x) = (1/2)sqrt(36 - (x - 3)²) on the domain 3 <= x < 9. To invert, set y = f(x) and make x the subject. Step 1: write the equation and clear the fraction. y = (1/2)sqrt(36 - (x - 3)²) 2y = sqrt(36 - (x - 3)²) Step 2: square both sides (legitimate here because both sides are >= 0 on this domain). 4y² = 36 - (x - 3)² Step 3: isolate the squared bracket. (x - 3)² = 36 - 4y² Step 4: take the square root. This is the step that needs care, because a square root gives two possibilities: x - 3 = +sqrt(36 - 4y²) or x - 3 = -sqrt(36 - 4y²). Factor 4 out of the surd: sqrt(36 - 4y²) = sqrt(4(9 - y²)) = 2 sqrt(9 - y²). So x = 3 + 2 sqrt(9 - y²) or x = 3 - 2 sqrt(9 - y²). Step 5: choose the correct branch using the domain. The domain of f is 3 <= x < 9, so x - 3 >= 0. The negative branch would give x - 3 <= 0, i.e. x <= 3, which is rejected (it only meets the domain at the single point x = 3). So take the positive branch: x = 3 + 2 sqrt(9 - y²). Step 6: rename the variable, since an inverse function is written in terms of x. f^{-1}(x) = 3 + 2 sqrt(9 - x²). (Its domain is the range of f, namely 0 < x <= 3.)