5. Graphical Applications of Derivatives

Finding Global Extrema

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Multiple Choice

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.
$y=x+\frac{2}{x};[0.25,3]$

Global maximum at $\left(0.25,8.25\right)$ , Global minimum at $\left(0,2\right)$

Global maximum at $\left(0.25,8.25\right)$ , Global minimum at $\left(\sqrt2,2\sqrt2\right)$

Global maximum at $\left(0.25,8.25\right)$ , Global minimum at $\left(3,3.67\right)$

Global maximum at $\left(\sqrt2,2\sqrt2\right)$ , Global minimum at $\left(-\sqrt2,-2\sqrt2\right)$

Verified step by step guidance

Identify the function to analyze: $ y = x + \frac{2}{x} $ and the interval $[0.25, 3]$.

Find the derivative of the function $ y = x + \frac{2}{x} $ to determine critical points. The derivative is $ y' = 1 - \frac{2}{x^2} $.

Set the derivative equal to zero to find critical points: $ 1 - \frac{2}{x^2} = 0 $. Solve for $ x $ to find the critical points within the interval.

Evaluate the function $ y = x + \frac{2}{x} $ at the critical points and at the endpoints of the interval $ x = 0.25 $ and $ x = 3 $.

Compare the function values obtained in the previous step to determine the global maximum and minimum values, and state them as ordered pairs.

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Expert video explanations

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.y=x+2x;[0.25,3]y=x+\frac{2}{x};[0.25,3]y=x+x2;[0.25,3]