Question

find ( f(0), f(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(0) )?

( f(0)=32 )

what is ( f(0) )?

( f(0)= )

Explanation:

Step1: Find the first - derivative

Using the chain rule \((u^n)^\prime = nu^{n - 1}u^\prime\), where \(u = 4x + 4\), \(n = 2\).
\(u^\prime=4\).
So \(f^\prime(x)=2(4x + 4)\times4=32x + 32\).

Step2: Find the second - derivative

Differentiate \(f^\prime(x)=32x + 32\) with respect to \(x\).
\(f^{\prime\prime}(x)=\frac{d}{dx}(32x + 32)=32\).
Since the second - derivative is a constant function, when \(x = 0\), \(f^{\prime\prime}(0)=32\).

Answer:

\(f^{\prime\prime}(0)=32\)