Question

Finding Parameter Values What values of \(a\) and \(b\) make \(f(x)=x^{3}+a x^{2}+b x\) have (a) a local maximum at \(x=-1\) and a local minimum at \(x=3 ?\) (b) a local minimum at \(x=4\) and a point of inflection at \(x=1 ?\)

Short answer

The parameter values that satisfy the conditions for part (a) are \(a=-4\) and \(b=-5\), and for part (b) are \(a=-3\) and \(b=-32\).

Step-by-step solution

Compute the derivative of the function

Find the first derivative of the function \(f(x)=x^{3}+a x^{2}+b x\) which yields \(f'(x) = 3x^{2} + 2ax + b\).

Apply the conditions for part (a)

Let's use the fact that f'(x) is equal to zero at local minima and maxima. So, we equate \(f'(x)\) to zero for \(x=-1\) and \(x=3\). This gives the system of equations: \[3(-1)^{2} + 2a(-1) + b = 0\]\[3(3)^{2} + 2a(3) + b = 0\] Simplify these equations to get: \[3 - 2a + b = 0\]\[27 + 6a + b = 0\]

Solve the system of equations for part (a)

Solving these equations for \(a\) and \(b\), will give: \(a=-4\) and \(b=-5\).

Apply the conditions for part (b)

In part b, the function has local minimum at \(x=4\), which means \(f'(4) = 0\). An inflection point at \(x=1\) means \(f''(1) = 0\). The second derivative \(f''(x) = 6x + 2a\), gives the system of equations: \[48 + 8a + b = 0\]\[6 + 2a = 0\]

Solve the system of equations for part (b)

Solving these equations for \(a\) and \(b\) yields: \(a=-3\) and \(b=-32\).