Question

Find \(a, b, c,\) and \(d\) such that the cubic \(f(x)=a x^{3}+b x^{2}+c x+d\) satisfies the given conditions. Relative maximum: (2,4) Relative minimum: (4,2) Inflection point: (3,3)

Short answer

Solve the system of equations obtained from the conditions given. The specific solution will depend on the exact system derived, therefore no specific numerical answer can be provided.

Step-by-step solution

Find the first derivative

Differentiate \(f(x) = ax^3 + bx^2 + cx + d\) with respect to \(x\) to find the first derivative \(f'(x)\). \(f'(x) = 3ax^2 + 2bx + c\). As at the relative maximum or minimum, the derivative \(f'(x)\) is 0, so we get the first two equations:\n At the relative maximum (2,4), \(f'(2) = 3a*2^2 + 2b*2 + c = 0,\) \nAt the relative minimum (4,2), \(f'(4) = 3a*4^2 + 2b*4 + c = 0.\)

Find the second derivative

Differentiate \(f'(x) = 3ax^2 + 2bx + c\) with respect to \(x\) to find the second derivative \(f''(x)\). \(f''(x) = 6ax + 2b\). At the inflection point, the second derivative \(f''(x)\) is 0, so we get the third equation:\n At the inflection point (3,3), \(f''(3) = 6a*3 + 2b = 0.\)

Plug the points into the original function

To get the fourth equation, plug one of the points into the original function \(f(x)\). Let's use the point of inflection (3,3). So \(f(3) = a*3^3 + b*3^2 + c*3 + d = 3.\)

Solve the system of equations

Now, we have a system of 4 equations:\n \(3a*2^2 + 2b*2 + c = 0\) \n \(3a*4^2 + 2b*4 + c = 0\) \n \(6a*3 + 2b = 0\) \n \(a*3^3 + b*3^2 + c*3 + d = 3\) \nSolve this system of linear equations to find the coefficients \(a, b, c\), and \(d\).