find the absolute maximum and minimum, if either exists, for ( f(x)=x+\frac{25}{x} ).\nfind the second…

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\neq0) 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.