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 ( square ) at ( x=square ).
b. there is no absolute maximum.

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 \(f^\prime(x)=1-25x^{-2}=1-\frac{25}{x^{2}}=\frac{x^{2}-25}{x^{2}}=\frac{(x - 5)(x + 5)}{x^{2}}\).
Set \(f^\prime(x)=0\), then \(\frac{(x - 5)(x + 5)}{x^{2}}=0\). The critical points are \(x=-5\) and \(x = 5\) (since \(x
eq0\) as the function is not defined at \(x = 0\)).

Step2: Use the second - derivative test

We know \(f^{\prime\prime}(x)=\frac{50}{x^{3}}\).

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

Step3: Analyze the behavior of the function 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:

B. There is no absolute maximum.