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To determine the range of the function [tex]\( f(x) = \frac{2x}{x^2 - 1} \)[/tex], we need to analyze its behavior and critical points. Here is a detailed step-by-step solution:
1. Identify the function's behavior and potential critical points:
The function [tex]\( f(x) = \frac{2x}{x^2 - 1} \)[/tex], where the denominator [tex]\( x^2 - 1 \)[/tex] can be factored as [tex]\( (x - 1)(x + 1) \)[/tex]. This function has vertical asymptotes at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], where the denominator becomes zero and the function is undefined.
2. Find the critical points by taking the derivative:
To locate the critical points, we need to determine where the derivative of the function is zero or undefined.
The derivative [tex]\( f'(x) \)[/tex] is found using the quotient rule:
[tex]\[ f'(x) = \frac{(2)(x^2 - 1) - (2x)(2x)}{(x^2 - 1)^2} = \frac{2(x^2 - 1) - 4x^2}{(x^2 - 1)^2} = \frac{2x^2 - 2 - 4x^2}{(x^2 - 1)^2} = \frac{-2x^2 - 2}{(x^2 - 1)^2} = \frac{-2(x^2 + 1)}{(x^2 - 1)^2} \][/tex]
Setting the numerator equal to zero to find the critical points:
[tex]\[ -2(x^2 + 1) = 0 \implies x^2 + 1 = 0 \implies x^2 = -1 \][/tex]
This equation has no real solutions since [tex]\( x^2 = -1 \)[/tex] implies [tex]\( x = \pm i \)[/tex], which are complex numbers. Hence, there are no real critical points where the derivative is zero.
3. Evaluate the function's behavior at the vertical asymptotes and limits:
We need to evaluate the behavior of the function as [tex]\( x \)[/tex] approaches the vertical asymptotes and towards positive and negative infinity.
- As [tex]\( x \to 1^+ \)[/tex] and [tex]\( x \to 1^- \)[/tex], the function [tex]\( f(x) \to \infty \)[/tex] and [tex]\( f(x) \to -\infty \)[/tex] respectively.
- As [tex]\( x \to -1^+ \)[/tex] and [tex]\( x \to -1^- \)[/tex], the function [tex]\( f(x) \to -\infty \)[/tex] and [tex]\( f(x) \to \infty \)[/tex] respectively.
4. Evaluate the limits at infinity:
- [tex]\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2x}{x^2 - 1} = \lim_{x \to \infty} \frac{2}{x - \frac{1}{x}} = 0\)[/tex]
- [tex]\(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2x}{x^2 - 1} = \lim_{x \to -\infty} \frac{2}{x - \frac{1}{x}} = 0\)[/tex]
5. Analyze the range from the behavior of the function:
Given that the limits at positive and negative infinity are both 0 and observing the function's behavior around the asymptotes, we find that the function [tex]\( f(x) \)[/tex] can theoretically take any real value as it spans from [tex]\( -\infty \)[/tex] to [tex]\( \infty \)[/tex]. Thus, the function does not have a bounded range in the real number system.
Since there are complex critical points [tex]\( \pm i \)[/tex] and the function approaches infinity near the asymptotes, the analysis shows no specific bounded real range values. So, when considering the limits and the asymptotic behaviors towards infinities, the value of [tex]\( (b - a) = 0 \)[/tex].
Therefore, the value of [tex]\( (b - a) \)[/tex] is:
[tex]\[ \boxed{0} \][/tex]
1. Identify the function's behavior and potential critical points:
The function [tex]\( f(x) = \frac{2x}{x^2 - 1} \)[/tex], where the denominator [tex]\( x^2 - 1 \)[/tex] can be factored as [tex]\( (x - 1)(x + 1) \)[/tex]. This function has vertical asymptotes at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], where the denominator becomes zero and the function is undefined.
2. Find the critical points by taking the derivative:
To locate the critical points, we need to determine where the derivative of the function is zero or undefined.
The derivative [tex]\( f'(x) \)[/tex] is found using the quotient rule:
[tex]\[ f'(x) = \frac{(2)(x^2 - 1) - (2x)(2x)}{(x^2 - 1)^2} = \frac{2(x^2 - 1) - 4x^2}{(x^2 - 1)^2} = \frac{2x^2 - 2 - 4x^2}{(x^2 - 1)^2} = \frac{-2x^2 - 2}{(x^2 - 1)^2} = \frac{-2(x^2 + 1)}{(x^2 - 1)^2} \][/tex]
Setting the numerator equal to zero to find the critical points:
[tex]\[ -2(x^2 + 1) = 0 \implies x^2 + 1 = 0 \implies x^2 = -1 \][/tex]
This equation has no real solutions since [tex]\( x^2 = -1 \)[/tex] implies [tex]\( x = \pm i \)[/tex], which are complex numbers. Hence, there are no real critical points where the derivative is zero.
3. Evaluate the function's behavior at the vertical asymptotes and limits:
We need to evaluate the behavior of the function as [tex]\( x \)[/tex] approaches the vertical asymptotes and towards positive and negative infinity.
- As [tex]\( x \to 1^+ \)[/tex] and [tex]\( x \to 1^- \)[/tex], the function [tex]\( f(x) \to \infty \)[/tex] and [tex]\( f(x) \to -\infty \)[/tex] respectively.
- As [tex]\( x \to -1^+ \)[/tex] and [tex]\( x \to -1^- \)[/tex], the function [tex]\( f(x) \to -\infty \)[/tex] and [tex]\( f(x) \to \infty \)[/tex] respectively.
4. Evaluate the limits at infinity:
- [tex]\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2x}{x^2 - 1} = \lim_{x \to \infty} \frac{2}{x - \frac{1}{x}} = 0\)[/tex]
- [tex]\(\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2x}{x^2 - 1} = \lim_{x \to -\infty} \frac{2}{x - \frac{1}{x}} = 0\)[/tex]
5. Analyze the range from the behavior of the function:
Given that the limits at positive and negative infinity are both 0 and observing the function's behavior around the asymptotes, we find that the function [tex]\( f(x) \)[/tex] can theoretically take any real value as it spans from [tex]\( -\infty \)[/tex] to [tex]\( \infty \)[/tex]. Thus, the function does not have a bounded range in the real number system.
Since there are complex critical points [tex]\( \pm i \)[/tex] and the function approaches infinity near the asymptotes, the analysis shows no specific bounded real range values. So, when considering the limits and the asymptotic behaviors towards infinities, the value of [tex]\( (b - a) = 0 \)[/tex].
Therefore, the value of [tex]\( (b - a) \)[/tex] is:
[tex]\[ \boxed{0} \][/tex]
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