Question

find the absolute maximum and minimum, if either exists, for $f(x)=x+\frac{49}{x}$. find the second derivative of $f$. $f(x)=\square$ select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute maximum is $\square$ at $x=\square$. b. there is no absolute maximum. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute minimum is $\square$ at $x=\square$. b. there is no absolute minimum.

Explanation:

Step1: Find the first derivative

Using the power rule, if \(y = x+\frac{49}{x}=x + 49x^{-1}\), then \(f^{\prime}(x)=1-49x^{-2}=1-\frac{49}{x^{2}}=\frac{x^{2}-49}{x^{2}}=\frac{(x - 7)(x + 7)}{x^{2}}\)

Step2: Find the second derivative

Differentiate \(f^{\prime}(x)\) again. Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u=x^{2}-49\), \(u^\prime = 2x\), \(v=x^{2}\), \(v^\prime=2x\). Then \(f^{\prime\prime}(x)=\frac{2x\cdot x^{2}-(x^{2}-49)\cdot2x}{x^{4}}=\frac{2x^{3}-2x^{3}+98x}{x^{4}}=\frac{98}{x^{3}}\)

Step3: Find critical points

Set \(f^{\prime}(x) = 0\), \(\frac{(x - 7)(x + 7)}{x^{2}}=0\). Since \(x
eq0\) (because \(f(x)\) is not defined at \(x = 0\)), then \(x-7=0\) or \(x + 7=0\), so \(x=7\) or \(x=-7\)

Step4: Use the second - derivative test

For \(x = 7\), \(f^{\prime\prime}(7)=\frac{98}{7^{3}}=\frac{98}{343}=\frac{2}{7}>0\). So \(f(x)\) has a local (and in this case, since we consider the domain \(x
eq0\) and the behavior of the function as \(x\to\pm\infty\), \(y=x+\frac{49}{x}\to\pm\infty\) as \(x\to\pm\infty\)) absolute minimum at \(x = 7\). \(f(7)=7+\frac{49}{7}=7 + 7=14\)
For \(x=-7\), \(f^{\prime\prime}(-7)=\frac{98}{(-7)^{3}}=-\frac{2}{7}<0\). So \(f(x)\) has a local maximum at \(x=-7\), \(f(-7)=-7+\frac{49}{-7}=-7-7=-14\). But as \(x\to\pm\infty\), \(y=x+\frac{49}{x}\to\pm\infty\), so there is no absolute maximum.

Answer:

\(f^{\prime}(x)=\frac{x^{2}-49}{x^{2}}\); A. The absolute maximum is \(\text{does not exist}\); A. The absolute minimum is at \(x = 7\)