Question

find ( f^{prime}(0), f^{prime prime}(0) ), and and determine whether ( f ) has a local minimum, local maximum, or neither at ( x = 0 ).

( f(x)=(4 x+4)^{2} )

what is ( f^{prime}(0) )?

( f^{prime}(0)= )

Explanation:

Step1: Differentiate \(f(x)\) using the chain rule

Let \(u = 4x + 4\), then \(f(x)=u^{2}\).
The chain rule states that \(\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}\).
\(\frac{df}{du} = 2u\) and \(\frac{du}{dx}=4\).
So \(f^{\prime}(x)=2(4x + 4)\cdot4=8(4x + 4)=32x+32\).

Step2: Evaluate \(f^{\prime}(0)\)

Substitute \(x = 0\) into \(f^{\prime}(x)\):
\(f^{\prime}(0)=32\times0 + 32\).

Answer:

\(32\)