Question

find the absolute maximum and minimum, if either exists, for ( f(x)=x+\frac{25}{x} ).
find the second derivative of ( f ).
( f^{prime prime}(x)=\frac{50}{x^{3}} )
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum is at ( x= ).
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 at ( x= ).
b. there is no absolute minimum.

Explanation:

Step1: Find the first derivative

The function is \(f(x)=x + \frac{25}{x}=x+25x^{-1}\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), the first derivative \(f^\prime(x)=1-25x^{-2}=1-\frac{25}{x^{2}}=\frac{x^{2}-25}{x^{2}}=\frac{(x - 5)(x + 5)}{x^{2}}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(\frac{(x - 5)(x + 5)}{x^{2}}=0\).
Since \(x
eq0\) (because \(f(x)\) is not defined at \(x = 0\)), we have \(x-5=0\) or \(x + 5=0\), so \(x=5\) or \(x=-5\).

Step3: Use the second - derivative test

The second derivative \(f^{\prime\prime}(x)=\frac{50}{x^{3}}\).

• When \(x = 5\), \(f^{\prime\prime}(5)=\frac{50}{5^{3}}=\frac{50}{125}=\frac{2}{5}>0\). So \(f(x)\) has a local minimum at \(x = 5\).

\(f(5)=5+\frac{25}{5}=5 + 5=10\).

• When \(x=-5\), \(f^{\prime\prime}(-5)=\frac{50}{(-5)^{3}}=\frac{50}{-125}=-\frac{2}{5}<0\). So \(f(x)\) has a local maximum at \(x=-5\).

\(f(-5)=-5+\frac{25}{-5}=-5-5=-10\).

Step4: Analyze the behavior as \(x\to\pm\infty\)

As \(x\to\infty\), \(f(x)=x+\frac{25}{x}\to\infty\) (since \(\lim_{x\to\infty}\frac{25}{x}=0\) and \(\lim_{x\to\infty}x=\infty\)).
As \(x\to-\infty\), \(f(x)=x+\frac{25}{x}\to-\infty\) (since \(\lim_{x\to-\infty}\frac{25}{x}=0\) and \(\lim_{x\to-\infty}x=-\infty\)).

Answer:

• For the absolute maximum: B. There is no absolute maximum.
• For the absolute minimum: A. The absolute minimum is \(10\) at \(x = 5\).