6. Graphical Applications of Derivatives

Finding Global Extrema

6. 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

Step 1: Begin by identifying the function and the interval provided. The function is y = x + \frac{2}{x}, and the interval is [0.25, 3]. The goal is to find the global maximum and minimum values of the function within this interval.

Step 2: Compute the derivative of the function to find critical points. The derivative of y = x + \frac{2}{x} is \frac{dy}{dx} = 1 - \frac{2}{x^2}. Set \frac{dy}{dx} = 0 to solve for x values where the slope is zero.

Step 3: Solve the equation 1 - \frac{2}{x^2} = 0. Rearrange to get \frac{2}{x^2} = 1, and then solve for x^2 = 2. This gives x = \sqrt{2} and x = -\sqrt{2}. Since the interval is [0.25, 3], only x = \sqrt{2} is relevant.

Step 4: Evaluate the function y = x + \frac{2}{x} at the critical point x = \sqrt{2}, as well as at the endpoints of the interval, x = 0.25 and x = 3. Substitute these values into the function to find the corresponding y values.

Step 5: Compare the y values obtained at the critical point and the endpoints to determine the global maximum and minimum. The global maximum occurs at the point with the highest y value, and the global minimum occurs at the point with the lowest y value.