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Determine f′(−2) for each function. a) f(x)=(x2−2x)(3x+1)

User Perocat
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1 Answer

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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:

User Euphoric
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