Question

find the extreme values of the function on the given interval.
$f(x)=x+\frac{6}{x}$ on $2,3$
enter the maximum value here, or enter none if there isnt one.
enter the minimum value here, or enter none if there isnt one.

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 \(1-\frac{6}{x^{2}}=0\).

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Since the interval is \([2,3]\), we consider \(x = \sqrt{6}\approx2.45\) (because \(-\sqrt{6}
otin[2,3]\)).

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 \(2\sqrt{6}\).