Analyzing the Given Function

The function we are given is \(f(x)=\sqrt{2-x}+\sqrt{2+x}\). We need to find its greatest value.

To understand the function, let's first determine its domain.

Finding the Domain of the Function

For the function \(f(x)=\sqrt{2-x}+\sqrt{2+x}\) to be defined, the expressions under the square roots must be non-negative.

• For \(\sqrt{2-x}\) to be defined, \(2-x \ge 0\), which means \(x \le 2\).
• For \(\sqrt{2+x}\) to be defined, \(2+x \ge 0\), which means \(x \ge -2\).

Both conditions must be met simultaneously, so the domain of the function is \([-2, 2]\).

Determining the Greatest Value

To find the greatest value of the function \(f(x)\) within its domain \([-2, 2]\), we can analyze the function's behavior. One way is to square the function, as \(f(x)\) is always non-negative (sum of square roots of non-negative numbers), and maximizing \(f(x)\) is equivalent to maximizing \(f(x)^2\).

Let's square \(f(x)\):

\(f(x)^2 = (\sqrt{2-x}+\sqrt{2+x})^2\)

Using the formula \((a+b)^2 = a^2 + b^2 + 2ab\):

\(f(x)^2 = (\sqrt{2-x})^2 + (\sqrt{2+x})^2 + 2(\sqrt{2-x})(\sqrt{2+x})\)

\(f(x)^2 = (2-x) + (2+x) + 2\sqrt{(2-x)(2+x)}\)

Simplifying the terms:

\(f(x)^2 = 2 - x + 2 + x + 2\sqrt{4 - x^2}\)

\(f(x)^2 = 4 + 2\sqrt{4 - x^2}\)

Now, to maximize \(f(x)^2\), we need to maximize the term \(2\sqrt{4 - x^2}\). This occurs when the expression inside the square root, \(4 - x^2\), is maximized.

The expression \(4 - x^2\) is maximized when \(x^2\) is minimized. Within the domain \([-2, 2]\), the minimum value of \(x^2\) is 0, which happens at \(x=0\).

Evaluating the Function at Critical Points

The analysis of \(f(x)^2\) suggests that a potential maximum occurs at \(x=0\). We also need to check the endpoints of the domain, \(x=-2\) and \(x=2\).

• At \(x=0\):
  \(f(0) = \sqrt{2-0} + \sqrt{2+0} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}\).
  We can write \(2\sqrt{2}\) as \(\sqrt{4} \times \sqrt{2} = \sqrt{4 \times 2} = \sqrt{8}\).
• At \(x=-2\):
  \(f(-2) = \sqrt{2-(-2)} + \sqrt{2+(-2)} = \sqrt{2+2} + \sqrt{2-2} = \sqrt{4} + \sqrt{0} = 2 + 0 = 2\).
• At \(x=2\):
  \(f(2) = \sqrt{2-2} + \sqrt{2+2} = \sqrt{0} + \sqrt{4} = 0 + 2 = 2\).

Comparing the Values

We found the following values for \(f(x)\) at the points of interest:

• \(f(0) = \sqrt{8}\)
• \(f(-2) = 2 = \sqrt{4}\)
• \(f(2) = 2 = \sqrt{4}\)

Comparing \(\sqrt{8}\) and \(\sqrt{4}\), it is clear that \(\sqrt{8} > \sqrt{4}\). Therefore, the greatest value of the function \(f(x)\) within its domain is \(\sqrt{8}\).

Conclusion

The greatest value of the function \(f(x)=\sqrt{2-x}+\sqrt{2+x}\) is \(\sqrt{8}\).

Revision Table: Key Concepts for Function Analysis

Additional Information: Alternative Methods for Finding Maximum Value

Besides squaring the function, another common method to find the maximum or minimum value of a differentiable function on a closed interval is using calculus:

• Find the derivative of the function, \(f'(x)\).
• Find critical points by setting \(f'(x) = 0\) and solving for \(x\). Also, consider points where \(f'(x)\) is undefined within the domain.
• Evaluate the function \(f(x)\) at all critical points that lie within the domain and at the endpoints of the domain.
• The largest of these values is the absolute maximum value of the function on the interval.

For \(f(x)=\sqrt{2-x}+\sqrt{2+x}\):

\(f'(x) = \frac{d}{dx}(\sqrt{2-x}) + \frac{d}{dx}(\sqrt{2+x})\)

\(f'(x) = \frac{1}{2\sqrt{2-x}}(-1) + \frac{1}{2\sqrt{2+x}}(1)\)

\(f'(x) = -\frac{1}{2\sqrt{2-x}} + \frac{1}{2\sqrt{2+x}}\)

Set \(f'(x) = 0\):

\(\frac{1}{2\sqrt{2+x}} = \frac{1}{2\sqrt{2-x}}\)

\(\sqrt{2+x} = \sqrt{2-x}\)

Squaring both sides:

\(2+x = 2-x\)

\(2x = 0\)

\(x = 0\)

The critical point \(x=0\) is within the domain \([-2, 2]\). Evaluating at \(x=0, -2, 2\) gives the same results as shown in the main solution, confirming \(\sqrt{8}\) as the greatest value.