Determine the intervals for which the function is concave up or concave down. State the inflection points.
$f(x)=4\(\ln\)(3x^2)$

• A. Concave down: $\(\left\)(-\(\infty\),0\(\right\))$; Concave up: $\(\left\)(0,\(\infty\]\right\))$; Inflection pt: $\(\left\)(0,0\(\right\))$
• B. Concave down: $\(\left\)(-\(\infty\),0\(\right\))$ & $\(\left\)(0,\(\infty\]\right\))$; No inflection pt
• C. Concave up: $\(\left\)(-\(\infty\),0\(\right\))$; Concave down: $\(\left\)(0,\(\infty\]\right\))$; Inflection pt: $\(\left\)(0,0\(\right\))$
• D. Concave up: $\(\left\)(-\(\infty\),0\(\right\))$; Concave down: $\(\left\)(0,\(\infty\]\right\))$; No inflection pt

Answer: (B)

Verified step by step guidance

1. To determine concavity, we need to find the second derivative of the function f(x) = 4ln(3x^2). Start by finding the first derivative f'(x).
2. The first derivative of f(x) = 4ln(3x^2) is found using the chain rule. First, differentiate the outer function: d/dx[ln(u)] = 1/u. Then, differentiate the inner function: d/dx[3x^2] = 6x. So, f'(x) = 4 * (1/(3x^2)) * 6x.
3. Simplify the expression for f'(x). This results in f'(x) = 8/x.
4. Now, find the second derivative f''(x) by differentiating f'(x) = 8/x. The derivative of 8/x is -8/x^2.
5. Analyze the sign of f''(x) = -8/x^2 to determine concavity. Since -8/x^2 is always negative for x ≠ 0, the function is concave down on both intervals (-∞, 0) and (0, ∞). There is no point where the concavity changes, so there are no inflection points.