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.
$f(x)=\frac{7}{2x-9}$

Concave down: $\left(-\infty,\infty\right)$ ; No Inflection Points

Concave up: $\left(-\infty,\infty\right)$ ; No Inflection Points

Concave down: $\left(-\infty,\frac92\right)$ ; Concave up: $\left(\frac92,\infty\right)$ ; Inflection point: $\left(\frac92,0\right)$

Concave down: $\left(-\infty,\frac92\right)$ ; Concave up: $\left(\frac92,\infty\right)$ ; No Inflection Points

Verified step by step guidance

Step 1: To determine concavity, calculate the second derivative of the function f(x). Start by finding the first derivative f'(x) using the given function f(x) = \frac{7}{2x-9}. Use the quotient rule or rewrite the function as 7(2x-9)^{-1} and apply the chain rule.

Step 2: Once you have f'(x), differentiate it again to find the second derivative f''(x). Simplify the expression for f''(x) as much as possible.

Step 3: Set the second derivative f''(x) equal to zero to find potential inflection points. Solve for x to determine where the concavity might change.

Step 4: Analyze the sign of f''(x) on the intervals determined by the critical points (from Step 3). If f''(x) > 0 on an interval, the function is concave up there. If f''(x) < 0, the function is concave down.

Step 5: Identify the intervals of concavity (concave up and concave down) and check if the second derivative changes sign at the critical points. If it does, those points are inflection points. Otherwise, there are no inflection points.