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$($\sqrt$2,2$\sqrt$2$\right$)$

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

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

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