Answer:
To determine the derivative of the function \(f(x) = (x^2 - 2x)(3x + 1)\) and evaluate it at \(x = -2\), you can use the product rule. The product rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by:
\[
(uv)' = u'v + uv'
\]
In this case, \(u(x) = x^2 - 2x\) and \(v(x) = 3x + 1\), so let's find their derivatives first:
1. \(u'(x) = \frac{d}{dx}(x^2 - 2x)\)
2. \(v'(x) = \frac{d}{dx}(3x + 1)\)
Now, let's calculate these derivatives:
1. \(u'(x) = 2x - 2\)
2. \(v'(x) = 3\)
Now, apply the product rule to find \(f'(x)\):
\[
f'(x) = (x^2 - 2x)(3) + (2x - 2)(3x + 1)
\]
Next, evaluate \(f'(-2)\):
\[
f'(-2) = (-2^2 - 2(-2))(3) + (2(-2) - 2)(3(-2) + 1)
\]
Now, compute the values:
\[
f'(-2) = (4 + 4)(3) + (-4 - 2)(-6 + 1)
\]
\[
f'(-2) = (8)(3) + (-6)(-5)
\]
\[
f'(-2) = 24 + 30
\]
\[
f'(-2) = 54
\]
So, \(f'(-2) = 54\) for the function \(f(x) = (x^2 - 2x)(3x + 1)\).
Explanation: