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. Step 1: Recall that concavity of a function is determined by the second derivative, f''(x). If f''(x) > 0 on an interval, the function is concave up on that interval. If f''(x) < 0 on an interval, the function is concave down on that interval. Inflection points occur where f''(x) changes sign.
2. Step 2: Start by finding the first derivative of the given function f(x) = 4ln(3x^2). Using the chain rule and the derivative of ln(u), we have f'(x) = 4 * (1/(3x^2)) * (6x) = 8/x.
3. Step 3: Next, find the second derivative f''(x) by differentiating f'(x) = 8/x. Using the power rule, rewrite f'(x) as 8x^(-1) and differentiate to get f''(x) = -8x^(-2) = -8/(x^2).
4. Step 4: Analyze the sign of f''(x) = -8/(x^2). Since x^2 is always positive for all x ≠ 0, the numerator -8 ensures that f''(x) is always negative. This means the function is concave down on both intervals (-∞, 0) and (0, ∞).
5. Step 5: Check for inflection points. Inflection points occur where f''(x) changes sign. Since f''(x) = -8/(x^2) does not change sign (it is always negative), there are no inflection points for this function.