For the function f(x) = x √(4 - x²) to be defined, the expression under the square root must be non-negative. Therefore, we set 4 - x² ≥ 0 which implies x ² ≤ 4. Taking the square root of both sides gives us the domain as [-2, 2].

For the function f(x) = x √(4 - x²) to be defined, the expression under the square root must be non-negative. Therefore, we set 4 - x² ≥ 0 which implies x ² ≤ 4. Taking the square root of both sides gives us the domain as [-2, 2].

Critical points occur where f'(x) = 0 or f'(x) is undefined. Setting the numerator of f'(x) to zero, we get 4 - 2x² = 0 which yields x ² = 2. Thus, x = √2 and x = -√2 are our critical points. The derivative is undefined when the denominator is zero, i.e., 4 - x² = 0 which means x = ± 2. These are the endpoints of the domain.

Critical points occur where f'(x) = 0 or f'(x) is undefined. Setting the numerator of f'(x) to zero, we get 4 - 2x² = 0 which yields x ² = 2. Thus, x = √2 and x = -√2 are our critical points. The derivative is undefined when the denominator is zero, i.e., 4 - x² = 0 which means x = ± 2. These are the endpoints of the domain.

Comparing the function values, we find that the maximum value is 2 at x = √2 and the minimum value is -2 at x = -√2. The values at the endpoints, f(2) = 0 and f(-2) = 0, are neither local maxima nor minima in this context, as the function is not defined beyond these points.

Comparing the function values, we find that the maximum value is 2 at x = √2 and the minimum value is -2 at x = -√2. The values at the endpoints, f(2) = 0 and f(-2) = 0, are neither local maxima nor minima in this context, as the function is not defined beyond these points.

Find the local maximum and minimum values of f(x) = x * sqrt(4 - $x^2$ ).

Answer: The local maximum value is 2 at x = $\sqrt{2}$ and the local minimum value is -2 at x = - $\sqrt{2}$ .

Step-by-step solution

Step 1: Determine the domain of the function

For the function f(x) = x $\sqrt{4 - x^2}$ to be defined, the expression under the square root must be non-negative. Therefore, we set 4 - $x^2 \ge 0$ which implies x $^2 \le 4$ . Taking the square root of both sides gives us the domain as [-2, 2].

Step 2: Find the first derivative of the function

To find the local maximum and minimum values, we need to find the critical points by calculating the first derivative of f(x) and setting it to zero. We use the product rule and the chain rule for differentiation. The product rule states that (uv)' = u'v + uv'. Here, $u = x$ and $v = \sqrt{4 - x^2}$ .

Step 3: Find the critical points

Critical points occur where f'(x) = 0 or f'(x) is undefined. Setting the numerator of f'(x) to zero, we get 4 - $2x^2 = 0$ which yields x $^2 = 2$ . Thus, $x = \sqrt{2}$ and $x = -\sqrt{2}$ are our critical points. The derivative is undefined when the denominator is zero, i.e., 4 - $x^2 = 0$ which means x $= \pm 2$ . These are the endpoints of the domain.

Step 4: Evaluate the function at critical points and endpoints

Now we evaluate the function f(x) at the critical points $x = \sqrt{2}$ and $x = -\sqrt{2}$ , as well as at the endpoints of the domain, $x = 2$ and $x = -2$ . This will help us identify the local maximum and minimum values.

Step 5: Identify local maximum and minimum values

Comparing the function values, we find that the maximum value is 2 at $x = \sqrt{2}$ and the minimum value is -2 at $x = -\sqrt{2}$ . The values at the endpoints, f(2) = 0 and f(-2) = 0, are neither local maxima nor minima in this context, as the function is not defined beyond these points.

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