5. Graphical Applications of Derivatives

Concavity

5. 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

To determine the concavity of the function, we need to find the second derivative of the function f(x) = $ \frac{7}{2x-9} $.

First, find the first derivative f'(x) using the quotient rule: $ f'(x) = \frac{d}{dx} \left( \frac{7}{2x-9} \right) $. The quotient rule states that if you have a function $ \frac{u}{v} $, then the derivative is $ \frac{u'v - uv'}{v^2} $.

Apply the quotient rule: Let u = 7 and v = 2x - 9. Then, u' = 0 and v' = 2. So, $ f'(x) = \frac{0 \cdot (2x-9) - 7 \cdot 2}{(2x-9)^2} = \frac{-14}{(2x-9)^2} $.

Next, find the second derivative f''(x) by differentiating f'(x): $ f''(x) = \frac{d}{dx} \left( \frac{-14}{(2x-9)^2} \right) $. Use the chain rule and power rule to differentiate.

Determine the intervals of concavity by setting f''(x) = 0 and solving for x. Analyze the sign of f''(x) on the intervals determined by the critical points to find where the function is concave up or concave down. Identify any inflection points where the concavity changes.