Question

find f(0), f(0), and and determine whether f has a local minimum, local maximum, or neither at x = 0.
f(x) = (4x + 4)^2
what is f(0)?
f(0) = 32
what is f(0)?
f(0) = 32
does the function have a local minimum, a local maximum, or neither?
a. the function has a local minimum at x = 0.
b. the function has a local maximum at x = 0.
c. the function has neither a local minimum nor a local maximum at x = 0.

Explanation:

Step1: Find the first - derivative

Use the chain rule \((u^n)^\prime=nu^{n - 1}u^\prime\). Let \(u = 4x + 4\), \(n = 2\). Then \(u^\prime=4\).
\(f^\prime(x)=2(4x + 4)\times4=32x+32\)
Substitute \(x = 0\) into \(f^\prime(x)\):
\(f^\prime(0)=32\times0 + 32=32\)

Step2: Find the second - derivative

Differentiate \(f^\prime(x)=32x + 32\) with respect to \(x\).
Since \((ax + b)^\prime=a\) (where \(a = 32\), \(b = 32\)), \(f^{\prime\prime}(x)=32\)
Substitute \(x = 0\) into \(f^{\prime\prime}(x)\): \(f^{\prime\prime}(0)=32\)

Step3: Use the second - derivative test

The second - derivative test states that if \(f^\prime(c)=0\) and \(f^{\prime\prime}(c)>0\), then \(f(x)\) has a local minimum at \(x = c\); if \(f^\prime(c)=0\) and \(f^{\prime\prime}(c)<0\), then \(f(x)\) has a local maximum at \(x = c\); if \(f^\prime(c)
eq0\), then \(x = c\) is not a local extremum.
Here \(f^\prime(0)=32
eq0\)

Answer:

\(f^\prime(0)=32\), \(f^{\prime\prime}(0)=32\), C. The function has neither a local minimum nor a local maximum at \(x = 0\).