Question

For the function \[ f(x) = x \sqrt{4 - x^2}, \] the maximum value in the range \(-2 \leq x \leq 2\) is ...................... (rounded off to two decimal places).

Hint:

For maximizing \(f(x)=x\sqrt{a^2-x^2}\), notice it represents half the equation of a circle. The maximum occurs at \(x=\pm \frac{a}{\sqrt{2}}\), and the maximum value is \(\frac{a^2}{2}\). Here \(a=2 \Rightarrow f_{\max} = 2\).

Answer 1

Correct Answer: 1.9

Step 1: Define the domain.
The function is \[ f(x) = x\sqrt{4 - x^2}. \] Since the square root requires \(4 - x^2 \geq 0\), we have domain \(-2 \leq x \leq 2\).
Step 2: Differentiate the function.
Let \[ f(x) = x (4 - x^2)^{1/2}. \] Using product rule: \[ f'(x) = (1) . (4 - x^2)^{1/2} + x . \frac{1}{2}(4 - x^2)^{-1/2} . (-2x). \] Simplify: \[ f'(x) = \sqrt{4 - x^2} - \frac{x^2}{\sqrt{4 - x^2}}. \]
Step 3: Set derivative = 0.
\[ f'(x) = \frac{(4 - x^2) - x^2}{\sqrt{4 - x^2}} = \frac{4 - 2x^2}{\sqrt{4 - x^2}}. \] So, \[ f'(x) = 0 \Rightarrow 4 - 2x^2 = 0 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}. \]
Step 4: Evaluate function at critical points and boundaries.
- At \(x = -2\): \[ f(-2) = -2 . \sqrt{4 - 4} = 0. \] - At \(x = 2\): \[ f(2) = 2 . \sqrt{4 - 4} = 0. \] - At \(x = \sqrt{2}\): \[ f(\sqrt{2}) = \sqrt{2} . \sqrt{4 - 2} = \sqrt{2} . \sqrt{2} = 2. \] - At \(x = -\sqrt{2}\): \[ f(-\sqrt{2}) = -\sqrt{2} . \sqrt{2} = -2. \]
Step 5: Choose the maximum.
The maximum value in \([-2,2]\) is: \[ f(x)_{\max} = 2. \] Rounded to two decimal places: \[ f(x)_{\max} = 2.00 \] Final Answer: \[ \boxed{2.00} \]