Question

consider the function $f(x)=\frac{3 + 7\ln x}{x}$ on the interval $(0,\infty)$. for each of the following, enter dne if a value does not exist. this function has an absolute minimum value equal to which is attained at $x =$. this function has an absolute maximum value equal to which is attained at $x =$.

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = 3 + 7\ln x\), \(u^\prime=\frac{7}{x}\), and \(v=x\), \(v^\prime = 1\). Then \(f^\prime(x)=\frac{\frac{7}{x}\cdot x-(3 + 7\ln x)\cdot1}{x^{2}}=\frac{7-(3 + 7\ln x)}{x^{2}}=\frac{4-7\ln x}{x^{2}}\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(\frac{4 - 7\ln x}{x^{2}}=0\). Since \(x^{2}>0\) for \(x\in(0,\infty)\), we solve \(4-7\ln x = 0\). Then \(\ln x=\frac{4}{7}\), and \(x = e^{\frac{4}{7}}\)

Step3: Use the second - derivative test

Find the second derivative. \(f^\prime(x)=\frac{4-7\ln x}{x^{2}}\), using the quotient rule again. Let \(u = 4-7\ln x\), \(u^\prime=-\frac{7}{x}\), \(v=x^{2}\), \(v^\prime = 2x\). Then \(f^{\prime\prime}(x)=\frac{-\frac{7}{x}\cdot x^{2}-(4 - 7\ln x)\cdot2x}{x^{4}}=\frac{-7x-8x + 14x\ln x}{x^{4}}=\frac{14\ln x-15}{x^{3}}\)
Substitute \(x = e^{\frac{4}{7}}\) into \(f^{\prime\prime}(x)\): \(f^{\prime\prime}(e^{\frac{4}{7}})=\frac{14\times\frac{4}{7}-15}{e^{\frac{12}{7}}}=\frac{8 - 15}{e^{\frac{12}{7}}}=-\frac{7}{e^{\frac{12}{7}}}<0\) (so \(x = e^{\frac{4}{7}}\) is a local maximum)
We also consider the behavior as \(x
ightarrow0^{+}\), \(\lim_{x
ightarrow0^{+}}f(x)=-\infty\) (since \(\ln x
ightarrow-\infty\) as \(x
ightarrow0^{+}\)) and as \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}\frac{3+7\ln x}{x}\). Using L'Hopital's rule (since \(\frac{\infty}{\infty}\) form), \(\lim_{x
ightarrow\infty}\frac{3 + 7\ln x}{x}=\lim_{x
ightarrow\infty}\frac{\frac{7}{x}}{1}=0\)

Answer:

The function has an absolute minimum value equal to \(-\infty\) (DNE in the context of real - valued minima for a non - bounded below function on \((0,\infty)\)) and an absolute maximum value equal to \(\frac{3 + 4}{e^{\frac{4}{7}}}=\frac{7}{e^{\frac{4}{7}}}\) which is attained at \(x = e^{\frac{4}{7}}\)