Question

consider the function \\(f(x) = \frac{\sqrt{x}}{108 + x^2}\\), \\(0 \le x \le 8\\).

the absolute maximum of \\(f(x)\\) (on the given interval) is at \\(x = \\)

and the absolute minimum of \\(f(x)\\) (on the given interval) is at \\(x = \\)

Explanation:

Find the derivative of the function

Using the Critical Points knowledge point

$$ f'(x) = \frac{\frac{1}{2\sqrt{x}}(108+x^2) - \sqrt{x}(2x)}{(108+x^2)^2} = \frac{108 - 3x^2}{2\sqrt{x}(108+x^2)^2} $$

Find the critical points in the interval

Using the Critical Points knowledge point

$$
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$$

Evaluate the function at critical points and endpoints

Using the Absolute Extrema on Closed Intervals knowledge point

$$
LATEXBLOCK1
$$

Determine the absolute extrema locations

Using the Absolute Extrema on Closed Intervals knowledge point

$$
LATEXBLOCK2
$$

Answer:

Consider the function \(f(x) = \frac{\sqrt{x}}{108 + x^2}\), \(0 \le x \le 8\).

The absolute maximum of \(f(x)\) (on the given interval) is at \(x =\) <blank>6</blank>

and the absolute minimum of \(f(x)\) (on the given interval) is at \(x =\) <blank>0</blank>