How do you determine the interval(s) on which the function #y= ln(x) / x^3# is concave up and concave down?

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Ezekiel Alexander · Accepted Answer

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Adam Adkins · Answer

undefined To determine the intervals where the function $y = \frac{\ln(x)}{x^3}$ is concave up and concave down, follow these steps: 1. Find the second derivative of the function. 2. Set the second derivative equal to zero and solve for $x$. These points are potential inflection points. 3. Create a sign chart using test points from each interval determined by the points found in step 2. 4. Determine the sign of the second derivative in each interval to determine concavity. Now, let's go through the steps: 1. The first derivative of $y = \frac{\ln(x)}{x^3}$ is: $$y' = \frac{1 - 3\ln(x)}{x^4}$$ 2. The second derivative can be found by differentiating $y'$: $$y'' = \frac{-3(1 - 4\ln(x))}{x^5}$$ 3. Set $y''$ equal to zero and solve for $x$: $$1 - 4\ln(x) = 0$$ $$4\ln(x) = 1$$ $$\ln(x) = \frac{1}{4}$$ $$x = e^{\frac{1}{4}}$$ 4. Create a sign chart using test points from each interval. Consider intervals $0 < x < e^{\frac{1}{4}}$ and $x > e^{\frac{1}{4}}$: - For $0 < x < e^{\frac{1}{4}}$, choose a test point such as $x = \frac{1}{2}$. Substituting this into $y''$, we get $y'' > 0$, so this interval is concave up. - For $x > e^{\frac{1}{4}}$, choose a test point such as $x = 2$. Substituting this into $y''$, we get $y'' < 0$, so this interval is concave down. Therefore, the function $y = \frac{\ln(x)}{x^3}$ is concave up on the interval $0 < x < e^{\frac{1}{4}}$ and concave down on the interval $x > e^{\frac{1}{4}}$.