Determination of Extreme Values for the Function f(x) arcsin(2x / (x^2 1))

In this article, we will explore the process of determining the extreme values of the function f(x) arcsin(2x / (x^2 1)). We will follow a systematic approach, starting with analyzing the domain of the function and then computing its derivative to find the critical points.

1. Determining the Domain

To find the domain of the function, we need to ensure that the input to the arcsin function lies within the interval [-1, 1]. We start by setting up the inequality:

1.1. Setting Up the Inequality

We need to solve the inequality:

-1 ≤ 2x / (x^2 1) ≤ 1

1.1.1. Right Side of the Inequality

For the right side, we have:

2x / (x^2 1) ≤ 1

Multiplying both sides by (x^2 1) (which is always positive) gives:

2x ≤ x^2 1

Rearranging the terms, we get:

x^2 - 2x 1 ≥ 0

This can be rewritten as:

(x - 1)^2 ≥ 0

Since (x - 1)^2 is always non-negative, this inequality holds for all real numbers x.

1.1.2. Left Side of the Inequality

For the left side, we have:

-1 ≤ 2x / (x^2 1)

Multiplying both sides by (x^2 1) gives:

- (x^2 1) ≤ 2x

Rearranging the terms, we get:

x^2 2x 1 ≥ 0

This can be rewritten as:

(x 1)^2 ≥ 0

Since (x 1)^2 is always non-negative, this inequality holds for all real numbers x.

Combining these results, we conclude that the domain of f(x) arcsin(2x / (x^2 1)) is all real numbers x ∈ ?.

2. Finding the Derivative and Critical Points

Next, we find the derivative of f(x) to determine the critical points and hence, the potential extreme values.

2.1. Computing the Derivative

Using the chain rule, the derivative of f(x) is given by:

f'(x) frac{1}{sqrt{1 - left(frac{2x}{x^2 1}right)^2}} cdot frac{d}{dx}left(frac{2x}{x^2 1}right)

We need to compute the derivative of 2x / (x^2 1) using the quotient rule. Let u 2x and v x^2 1, then:

u' 2, quad v' 2x

The quotient rule gives:

frac{d}{dx}left(frac{2x}{x^2 1}right) frac{u'v - uv'}{v^2} frac{2(x^2 1) - 2x(2x)}{(x^2 1)^2} frac{2x^2 2 - 4x^2}{(x^2 1)^2} frac{2 - 2x^2}{(x^2 1)^2}

Substituting this back into the derivative of f(x), we get:

f'(x) frac{1}{sqrt{1 - left(frac{2x}{x^2 1}right)^2}} cdot frac{2 - 2x^2}{(x^2 1)^2}

2.2. Setting the Derivative to Zero

To find the critical points, we set f'(x) to zero:

frac{2 - 2x^2}{(x^2 1)^2} 0

This implies:

2 - 2x^2 0 implies x^2 1 implies x pm 1

3. Evaluating the Function at Critical Points

Now we evaluate f(x) at the critical points x 1 and x -1:

3.1. Evaluating at x 1

For x 1:

f(1) arcsinleft(frac{2 cdot 1}{1^2 1}right) arcsinleft(frac{2}{2}right) arcsin(1) frac{pi}{2}

3.2. Evaluating at x -1

For x -1:

f(-1) arcsinleft(frac{2 cdot (-1)}{(-1)^2 1}right) arcsinleft(frac{-2}{2}right) arcsin(-1) -frac{pi}{2}

4. Conclusion

We have found that the extreme values of the function f(x) arcsin(2x / (x^2 1)) are:

- Maximum: f(1) frac{pi}{2}
- Minimum: f(-1) -frac{pi}{2}

Thus, the extreme values of the function are frac{pi}{2} and -frac{pi}{2}.