Question

determine all critical points for the following function.
f(x) = 2x(16 - x)^3
what is the derivative of f(x) = 2x(16 - x)^3?
f(x) = □

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 2x\) and \(v=(16 - x)^3\). First, find \(u^\prime\) and \(v^\prime\).
\(u^\prime=\frac{d}{dx}(2x)=2\).
To find \(v^\prime\), use the chain rule. If \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(g(x)=16 - x\) and \(f(g)=g^3\). Then \(g^\prime(x)=- 1\) and \(f^\prime(g) = 3g^2\). So \(v^\prime=\frac{d}{dx}(16 - x)^3=3(16 - x)^2\cdot(-1)=-3(16 - x)^2\).

Step2: Calculate \(f^\prime(x)\)

By the product rule \(f^\prime(x)=u^\prime v+uv^\prime\).
Substitute \(u = 2x\), \(u^\prime = 2\), \(v=(16 - x)^3\), and \(v^\prime=-3(16 - x)^2\) into the formula:

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Answer:

\(f^\prime(x)=8(16 - x)^2(4 - x)\)