find the extreme values of the function on the given interval.\n$f(x)=x+\\frac{6}{x}$ on $2,3$\nenter the…

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=x + \frac{6}{x}), and its derivative (f^\prime(x)) is found using the power rule. For (y = x^n), (y^\prime=nx^{n - 1}). So (f^\prime(x)=1-\frac{6}{x^{2}}).

Step2: Find the critical points

Set (f^\prime(x) = 0), then (1-\frac{6}{x^{2}}=0). [ \begin{align*} 1-\frac{6}{x^{2}}&=0\ \frac{x^{2}-6}{x^{2}}&=0\ x^{2}-6&=0\ x^{2}&=6\ x&=\pm\sqrt{6} \end{align*} ] Since the interval is ([2,3]), we consider (x = \sqrt{6}\approx2.45) (because (-\sqrt{6}\notin[2,3])).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate (f(x)) at (x = 2): (f(2)=2+\frac{6}{2}=2 + 3=5).
• Evaluate (f(x)) at (x=\sqrt{6}): (f(\sqrt{6})=\sqrt{6}+\frac{6}{\sqrt{6}}=\sqrt{6}+\sqrt{6}=2\sqrt{6}\approx4.9).
• Evaluate (f(x)) at (x = 3): (f(3)=3+\frac{6}{3}=3 + 2=5).

Answer:

The maximum value is (5) and the minimum value is (2\sqrt{6}).