A-Level Functions

Given g(f(x+y)) = f(x) + (2x + y)g(y) for all real x, y. Put y = -2x, so x + y = -x: g(f(-x)) = f(x) + (2x + (-2x))g(-2x) = f(x) + 0*g(-2x) = f(x). Hence g(f(-x)) = f(x) for all x. (*) Replacing x by -t in (*): g(f(t)) = f(-t) for all t. (**) Now apply (**) with t = x + y to the left side of the original identity: g(f(x+y)) = f(-(x+y)) = f(-x - y). But the original identity says g(f(x+y)) = f(x) + (2x + y)g(y). Therefore f(-x - y) = f(x) + (2x + y)g(y), as required.

"Algebraically" means you may not read the range off a sketch. The standard algebraic method is: put y = f(x), rearrange into a QUADRATIC in x, and then use the fact that x has to be a real number, so that quadratic's discriminant cannot be negative. The set of y values for which a real x exists is exactly the range. Step 1: Let y = f(x). Multiply both sides by the denominator to clear the fraction, then collect every term on one side so the equation reads (quadratic in x, with coefficients involving y) = 0. Step 2: Apply the reality condition. A quadratic ax² + bx + c = 0 has a real solution x if and only if b² - 4ac >= 0. Writing out b² - 4ac for the quadratic from Step 1 and simplifying gives the condition (4y + 1)/y >= 0. Step 3: Solve (4y + 1)/y >= 0. Do NOT multiply both sides by y, because the sign of y is unknown and multiplying by a negative number would flip the inequality. Multiply by y² instead, which is positive for every y other than 0, so the direction is safe: y(4y + 1) >= 0. The critical values are y = -1/4 and y = 0. Test the sign of the product y(4y + 1) in each region: y < -1/4: (negative)(negative) = positive -> satisfied -1/4 < y < 0: (negative)(positive) = negative -> not satisfied y > 0: (positive)(positive) = positive -> satisfied The boundary y = -1/4 is included, because there the product is 0, the discriminant is 0 and x is still real (a repeated root). So y <= -1/4 or y >= 0. Step 4: Remove y = 0. y = 0 appeared only as a critical value of the sign test; it is excluded from the original inequality (4y + 1)/y >= 0 because it makes the denominator zero, and no x maps to it. So the second branch is y > 0, not y >= 0. Step 5: Write the range in interval notation. R_f = (-infinity, -1/4] union (0, infinity).

Step 1. Write y for the output, then make the denominator something you can undo, by completing the square. y = 1/(x² + 6x + 5) Half of 6 is 3, so x² + 6x = (x + 3)² − 9. x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4 So y = 1/((x + 3)² − 4). Step 2. Take the reciprocal of both sides, which frees the bracket from the fraction. 1/y = (x + 3)² − 4 Step 3. Add 4 to both sides to leave the square on its own. (x + 3)² = 1/y + 4 Step 4. Square root. This normally gives two possibilities, x + 3 = +√(1/y + 4) or x + 3 = −√(1/y + 4). The domain of f is x ≥ −3, so x + 3 ≥ 0, which rules out the negative root. x + 3 = √(1/y + 4) x = −3 + √(1/y + 4) Step 5. The inverse function sends an output back to the input that produced it, so this expression for x in terms of y is exactly the rule for f⁻¹. Rename the variable y as x. f⁻¹(x) = −3 + √(1/x + 4)