A curve C has equation y = 3x^4 - 8x^3 - 3. Find dy/dx and d2y/dx2. Verify C has a stationary point at x = 2. Determine the nature of this stationary point, giving a reason for the answer.

y = 3x⁴ - 8x³ - 3Bringing the power down on each term and subtracting one from the power (as to differentiate, nx^m —> nmx^m-1), we get for each derivative:dy/dx = 12x³ - 24x²d²y/dx² = 36x² - 48x.This answers the first part.When a curve has a stationary point at a point X, f’(X) = 0 [note: f’(x) is another notation for dy/dx and f’’(x) for d²y/dx² and so on]. So, to verify that our function has a stationary point at x = 2, we substitute x = 2 into the first derivative. 12(2)³ - 24(2)² = 12(8) - 24(4) = 96 - 96 = 0. Hence, C has a stationary point at x = 2.To determine the nature of this stationary point, we substitute x = 2 into the second derivative. If f’’(x) < 0, the stationary point is a local maximum point of the function. If f’’(x) > 0, the stationary point is a local minimum point of the function. If f’’(x) = 0, it could be either.So, substituting in x = 2, 36(2)² - 48(2) = 36(4) - 48(2) = 144 - 96 = 48. 48 > 0, and so the stationary point is a local maximum.