Question

1. 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 )?

Explanation:

Step1: Find the derivative of \(f(x)\)

The derivative of \(f(x)=x^{3}+ax^{2}+bx\) is \(f^{\prime}(x)=3x^{2}+2ax + b\).
Since \(x = - 1\) and \(x = 3\) are critical points, \(f^{\prime}(-1)=0\) and \(f^{\prime}(3)=0\).
For \(x=-1\):
\(f^{\prime}(-1)=3\times(-1)^{2}+2a\times(-1)+b=0\), which simplifies to \(3 - 2a + b=0\), or \(b=2a - 3\).
For \(x = 3\):
\(f^{\prime}(3)=3\times3^{2}+2a\times3 + b=0\), which simplifies to \(27+6a + b=0\).

Step2: Substitute \(b = 2a - 3\) into \(27+6a + b=0\)

Substitute \(b\) into \(27+6a + b=0\):
\(27+6a+(2a - 3)=0\).
Combine like - terms: \(27-3+6a + 2a=0\), so \(24 + 8a=0\).
Solve for \(a\): \(8a=-24\), then \(a=-3\).

Step3: Find the value of \(b\)

Substitute \(a=-3\) into \(b = 2a - 3\).
\(b=2\times(-3)-3=-6 - 3=-9\).

Answer:

\(a=-3\) and \(b = - 9\)