determine all critical points for the following function.\nf(x) = 2x(16 - x)^3\nwhat is the derivative of…

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: [ \begin{align*} f^\prime(x)&=2\cdot(16 - x)^3+2x\cdot(-3)(16 - x)^2\ &=2(16 - x)^3-6x(16 - x)^2\ &=2(16 - x)^2[(16 - x)-3x]\ &=2(16 - x)^2(16 - 4x)\ &=8(16 - x)^2(4 - x) \end{align*} ]

Answer:

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