A-Level Differentiation: worked solution

Question

A cubic f(x) = ax^3 + bx^2 + cx + d has a turning point at x = -1 and passes through (2, 10) and (-2, -2). If also INT from -2 to 2 of f(x) dx = 16, determine a, b, c, d.

There are four unknowns, so translate each of the four conditions into an equation, then solve the system. f(x) = ax³ + bx² + cx + d, so f'(x) = 3ax² + 2bx + c. Condition 1 - a turning point at x = -1 means f'(-1) = 0: 3a(-1)² + 2b(-1) + c = 0 3a - 2b + c = 0 ... (1) Condition 2 - the curve passes through (2, 10), so f(2) = 10: 8a + 4b + 2c + d = 10 ... (2) Condition 3 - the curve passes through (-2, -2), so f(-2) = -2: -8a + 4b - 2c + d = -2 ... (3) Condition 4 - the definite integral: INT from -2 to 2 of f(x) dx = [ ax⁴/4 + bx³/3 + cx²/2 + dx ] from -2 to 2 The terms in x⁴ and x² are EVEN, so they take the same value at x = 2 and at x = -2 and cancel on subtraction. Only the odd-power terms survive: = (8b/3 + 2d) - (-8b/3 - 2d) = 16b/3 + 4d So 16b/3 + 4d = 16. Multiply through by 3/4: 4b + 3d = 12 ... (4) Now solve. Adding (2) and (3) eliminates both a and c, since their coefficients are opposite: (8a - 8a) + (4b + 4b) + (2c - 2c) + (d + d) = 10 + (-2) 8b + 2d = 8, so 4b + d = 4 ... (5) Subtract (5) from (4): 3d - d = 12 - 4, so 2d = 8 and d = 4 Substitute back into (5): 4b + 4 = 4, so b = 0 Subtracting (3) from (2) eliminates b and d instead: (8a + 8a) + (2c + 2c) = 10 - (-2) 16a + 4c = 12, so 4a + c = 3 ... (6) With b = 0, equation (1) becomes 3a + c = 0, so c = -3a Substitute into (6): 4a + (-3a) = 3, so a = 3 Then c = -3(3) = -9 Check: f(x) = 3x³ - 9x + 4 gives f(2) = 24 - 18 + 4 = 10, f(-2) = -24 + 18 + 4 = -2, and f'(-1) = 9 - 9 = 0. ✓ So a = 3, b = 0, c = -9, d = 4.