Question

find the x - coordinates of all relative minima of f(x).

f(x)=x^{3}+6x^{2}-63x - 32

answer attempt 1 out of 3

additional solution no solution

Explanation:

Step1: Find the first derivative

The derivative of \(f(x)=x^{3}+6x^{2}-63x - 32\) is \(f^\prime(x)=3x^{2}+12x - 63\).
Factor out 3: \(f^\prime(x)=3(x^{2}+4x - 21)\).
Factor the quadratic: \(f^\prime(x)=3(x + 7)(x - 3)\).

Step2: Find critical points

Set \(f^\prime(x)=0\).
\(3(x + 7)(x - 3)=0\).
Solutions are \(x=-7\) and \(x = 3\).

Step3: Use the second - derivative test

Find the second derivative \(f^{\prime\prime}(x)=6x+12\).
For \(x=-7\): \(f^{\prime\prime}(-7)=6\times(-7)+12=-42 + 12=-30<0\) (relative maximum).
For \(x = 3\): \(f^{\prime\prime}(3)=6\times3+12=18 + 12=30>0\) (relative minimum).

Answer:

\(3\)