Question

determine all critical points for the following function.

f(x)=2x(8 - x)^3

what is the derivative of f(x)=2x(8 - x)^3?

f(x)=\square

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 \(v=(8 - x)^{3}\). First, find \(u^\prime\) and \(v^\prime\).
\(u^\prime=\frac{d}{dx}(2x)=2\)
For \(v=(8 - x)^{3}\), use the chain rule. Let \(t = 8 - x\), then \(v=t^{3}\). \(\frac{dv}{dt}=3t^{2}\) and \(\frac{dt}{dx}=- 1\). So \(v^\prime=\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}=3(8 - x)^{2}\cdot(-1)=-3(8 - x)^{2}\)

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

By the product rule \(f^\prime(x)=u^\prime v+uv^\prime\)
\(f^\prime(x)=2(8 - x)^{3}+2x\cdot(-3)(8 - x)^{2}\)
Factor out \(2(8 - x)^{2}\)
\(f^\prime(x)=2(8 - x)^{2}[(8 - x)-3x]\)
Simplify the expression inside the brackets: \((8 - x)-3x=8 - 4x\)
So \(f^\prime(x)=2(8 - x)^{2}(8 - 4x)=8(8 - x)^{2}(2 - x)\)

Answer:

\(f^\prime(x)=8(8 - x)^{2}(2 - x)\)