6. Graphical Applications of Derivatives

Concavity

6. Graphical Applications of Derivatives

Concavity

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Multiple Choice

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

Concave down: $\left(-\infty,0\right)$ ; Concave up: $\left(0,\infty\right)$ ; Inflection pt: $\left(0,0\right)$

Concave down: $\left(-\infty,0\right)$ & $\left(0,\infty\right)$ ; No inflection pt

Concave up: $\left(-\infty,0\right)$ ; Concave down: $\left(0,\infty\right)$ ; Inflection pt: $\left(0,0\right)$

Concave up: $\left(-\infty,0\right)$ ; Concave down: $\left(0,\infty\right)$ ; No inflection pt

Verified step by step guidance

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.

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.

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).

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, ∞).

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.