find the absolute maximum and absolute minimum values of f on the given interval.\n\nf(x)=x + \\frac{25}{x}…

Explanation:

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

The function is (f(x)=x + \frac{25}{x}=x+25x^{-1}). Using the power rule ((x^n)^\prime=nx^{n - 1}), the 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\in[0.2,20]), we solve (x - 5=0) (because (x+5\neq0) for (x\in[0.2,20]) and (x^{2}\neq0) for (x\in[0.2,20])). So (x = 5) is a critical point.

Step3: Evaluate the function at the critical point and endpoints

• Evaluate (f(x)) at (x = 0.2): (f(0.2)=0.2+\frac{25}{0.2}=0.2 + 125=125.2).
• Evaluate (f(x)) at (x = 5): (f(5)=5+\frac{25}{5}=5 + 5=10).
• Evaluate (f(x)) at (x = 20): (f(20)=20+\frac{25}{20}=20+1.25=21.25).

Answer:

absolute minimum value: (10) absolute maximum value: (125.2)