To find the global maximum and global minimum of the function $f(x) = x \sqrt{1 - x^2}$ on the interval $[-2, 1]$, follow these steps:

Step 1: Check the domain of the function

The square root term $\sqrt{1 - x^2}$ is defined only when $1 - x^2 \geq 0$, which means: $-1 \leq x \leq 1$ So, $f(x)$ is undefined for $x \in [-2, -1)$. Therefore, we restrict our evaluation to $x \in [-1, 1]$.

Step 2: Derivative to find critical points

The derivative of $f(x)$ is: $f'(x) = \sqrt{1 - x^2} + x \cdot \frac{-2x}{2\sqrt{1 - x^2}}$ Simplify: $f'(x) = \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}}$ Combine under a common denominator: $f'(x) = \frac{(1 - x^2) - x^2}{\sqrt{1 - x^2}} = \frac{1 - 2x^2}{\sqrt{1 - x^2}}$

Set $f'(x) = 0$: $1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}$

Step 3: Evaluate $f(x)$ at critical points and endpoints

The critical points are $x = \pm \frac{\sqrt{2}}{2}$. The endpoints of the interval are $x = -1$ and $x = 1$.

1. $f\left(-1\right) = (-1) \sqrt{1 - (-1)^2} = (-1) \cdot 0 = 0$
2. $f\left(1\right) = (1) \sqrt{1 - (1)^2} = (1) \cdot 0 = 0$
3. $f\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} \sqrt{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\sqrt{2}}{2} \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2}$
4. $f\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\sqrt{2}}{2} \sqrt{1 - \left(-\frac{\sqrt{2}}{2}\right)^2} = -\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = -\frac{1}{2}$

Step 4: Determine the global extrema

The values of $f(x)$ are:

• At $x = -1$: $f(x) = 0$
• At $x = 1$: $f(x) = 0$
• At $x = \frac{\sqrt{2}}{2}$: $f(x) = \frac{1}{2}$
• At $x = -\frac{\sqrt{2}}{2}$: $f(x) = -\frac{1}{2}$

Thus:

• Global maximum: $f(x) = \frac{1}{2}$ at $x = \frac{\sqrt{2}}{2}$
• Global minimum: $f(x) = -\frac{1}{2}$ at $x = -\frac{\sqrt{2}}{2}$

Final Answer:

• Global max: $\frac{1}{2}$ at $x = \frac{\sqrt{2}}{2}$
• Global min: $-\frac{1}{2}$ at $x = -\frac{\sqrt{2}}{2}$

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Related Questions:

1. How to find the derivative of a function involving square roots?
2. What are the steps to determine critical points of a function?
3. Why does the function have a restricted domain?
4. Can this method apply to a similar trigonometric function?
5. How can global extrema be verified visually?

Tip: Always verify the domain of the function when dealing with square roots or logarithms.