studio di funzioni logaritmiche. studiare le seguenti funzioni fino alla derivata seconda, tracciarne il…

Explanation:

Step1: Find the first - derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{3}) and (v=\ln x - 1). (u^\prime=3x^{2}), (v^\prime=\frac{1}{x}). (f^\prime(x)=3x^{2}(\ln x - 1)+x^{3}\cdot\frac{1}{x}=3x^{2}\ln x-3x^{2}+x^{2}=3x^{2}\ln x - 2x^{2}=x^{2}(3\ln x - 2))

Step2: Find the critical points

Set (f^\prime(x) = 0). Since (x^{2}\gt0) for (x\gt0) (the domain of (y = \ln x) is (x\gt0)), we solve (3\ln x - 2 = 0). (3\ln x=2), (\ln x=\frac{2}{3}), (x = e^{\frac{2}{3}})

Step3: Find the second - derivative

Use the product rule again on (f^\prime(x)=x^{2}(3\ln x - 2)). Let (u = x^{2}), (u^\prime = 2x), (v = 3\ln x - 2), (v^\prime=\frac{3}{x}). (f^{\prime\prime}(x)=2x(3\ln x - 2)+x^{2}\cdot\frac{3}{x}=6x\ln x-4x + 3x=6x\ln x - x=x(6\ln x - 1))

Step4: Classify the critical point

Evaluate (f^{\prime\prime}(e^{\frac{2}{3}})): (f^{\prime\prime}(e^{\frac{2}{3}})=e^{\frac{2}{3}}(6\cdot\frac{2}{3}-1)=e^{\frac{2}{3}}(4 - 1)=3e^{\frac{2}{3}}\gt0) So (x = e^{\frac{2}{3}}) is a local minimum.

Step5: Find inflection points

Set (f^{\prime\prime}(x)=0). Since (x\gt0), we solve (6\ln x - 1 = 0). (6\ln x=1), (\ln x=\frac{1}{6}), (x = e^{\frac{1}{6}})

To draw the graph:

• As (x\to0^{+}), (f(x)\to+\infty) (because (x^{3}\to0^{+}) and (\ln x - 1\to-\infty) but the power - law function (x^{3}) dominates the growth of (\ln x - 1) near (x = 0)).
• The function has a local minimum at (x = e^{\frac{2}{3}}) and an inflection point at (x = e^{\frac{1}{6}}).

Answer:

Local minimum at (x = e^{\frac{2}{3}}), inflection point at (x = e^{\frac{1}{6}})