Answer :
Certainly! Let's analyze the function [tex]\( f(x) = \frac{x^2}{x^2 - 1} \)[/tex] step-by-step.
1. Setting Up the Function:
[tex]\[ f(x) = \frac{x^2}{x^2 - 1} \][/tex]
Here, [tex]\( f(x) \)[/tex] is a rational function where the numerator is [tex]\( x^2 \)[/tex] and the denominator is [tex]\( x^2 - 1 \)[/tex].
2. Identifying the Domain:
The function will be undefined where the denominator is zero. We need to determine where [tex]\( x^2 - 1 = 0 \)[/tex]:
[tex]\[ x^2 - 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex]. Thus, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
3. Simplifying the Fraction:
To further analyze the function, let's see if we can simplify it. However, in this particular case, [tex]\( \frac{x^2}{x^2 - 1} \)[/tex] does not simplify through common factors.
4. Asymptotic Behavior:
- Vertical Asymptotes:
Since the function becomes undefined at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], these points are the vertical asymptotes of [tex]\( f(x) \)[/tex].
- Horizontal Asymptotes:
To find horizontal asymptotes, we analyze the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^2}{x^2 - 1} \][/tex]
Dividing the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{1}{1 - \frac{1}{x^2}} = 1 \][/tex]
Therefore, the horizontal asymptote of the function is [tex]\( y = 1 \)[/tex].
5. Critical Points and Derivatives:
- First Derivative:
To find critical points, we will find the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{x^2}{x^2 - 1} \right) \][/tex]
Using the quotient rule [tex]\( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)[/tex], where [tex]\( u = x^2 \)[/tex] and [tex]\( v = x^2 - 1 \)[/tex]:
[tex]\[ u' = 2x, \quad v' = 2x \][/tex]
[tex]\[ f'(x) = \frac{(2x)(x^2 - 1) - (x^2)(2x)}{(x^2 - 1)^2} \][/tex]
[tex]\[ f'(x) = \frac{2x^3 - 2x - 2x^3}{(x^2 - 1)^2} = \frac{-2x}{(x^2 - 1)^2} \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ -2x = 0 \implies x = 0 \][/tex]
Thus, there is a critical point at [tex]\( x = 0 \)[/tex].
- Second Derivative:
For concavity, we can compute the second derivative, but as of now, we've identified key aspects of [tex]\( f(x) \)[/tex].
6. Behavior Around Critical Points and Asymptotes:
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2}{0^2 - 1} = 0 \][/tex]
Thus, the function passes through the point [tex]\( (0, 0) \)[/tex].
- Near the Vertical Asymptotes [tex]\( x = \pm 1 \)[/tex]:
The function approaches infinity as [tex]\( x \)[/tex] approaches [tex]\( 1 \)[/tex] and [tex]\( -1 \)[/tex].
In conclusion, [tex]\( f(x) = \frac{x^2}{x^2 - 1} \)[/tex] is defined for all real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], has a horizontal asymptote at [tex]\( y = 1 \)[/tex], vertical asymptotes at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], and passes through the origin [tex]\( (0, 0) \)[/tex].
1. Setting Up the Function:
[tex]\[ f(x) = \frac{x^2}{x^2 - 1} \][/tex]
Here, [tex]\( f(x) \)[/tex] is a rational function where the numerator is [tex]\( x^2 \)[/tex] and the denominator is [tex]\( x^2 - 1 \)[/tex].
2. Identifying the Domain:
The function will be undefined where the denominator is zero. We need to determine where [tex]\( x^2 - 1 = 0 \)[/tex]:
[tex]\[ x^2 - 1 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Therefore, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex]. Thus, the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
3. Simplifying the Fraction:
To further analyze the function, let's see if we can simplify it. However, in this particular case, [tex]\( \frac{x^2}{x^2 - 1} \)[/tex] does not simplify through common factors.
4. Asymptotic Behavior:
- Vertical Asymptotes:
Since the function becomes undefined at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], these points are the vertical asymptotes of [tex]\( f(x) \)[/tex].
- Horizontal Asymptotes:
To find horizontal asymptotes, we analyze the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{x^2}{x^2 - 1} \][/tex]
Dividing the numerator and the denominator by [tex]\( x^2 \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{1}{1 - \frac{1}{x^2}} = 1 \][/tex]
Therefore, the horizontal asymptote of the function is [tex]\( y = 1 \)[/tex].
5. Critical Points and Derivatives:
- First Derivative:
To find critical points, we will find the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{x^2}{x^2 - 1} \right) \][/tex]
Using the quotient rule [tex]\( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)[/tex], where [tex]\( u = x^2 \)[/tex] and [tex]\( v = x^2 - 1 \)[/tex]:
[tex]\[ u' = 2x, \quad v' = 2x \][/tex]
[tex]\[ f'(x) = \frac{(2x)(x^2 - 1) - (x^2)(2x)}{(x^2 - 1)^2} \][/tex]
[tex]\[ f'(x) = \frac{2x^3 - 2x - 2x^3}{(x^2 - 1)^2} = \frac{-2x}{(x^2 - 1)^2} \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex] to find critical points:
[tex]\[ -2x = 0 \implies x = 0 \][/tex]
Thus, there is a critical point at [tex]\( x = 0 \)[/tex].
- Second Derivative:
For concavity, we can compute the second derivative, but as of now, we've identified key aspects of [tex]\( f(x) \)[/tex].
6. Behavior Around Critical Points and Asymptotes:
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0^2}{0^2 - 1} = 0 \][/tex]
Thus, the function passes through the point [tex]\( (0, 0) \)[/tex].
- Near the Vertical Asymptotes [tex]\( x = \pm 1 \)[/tex]:
The function approaches infinity as [tex]\( x \)[/tex] approaches [tex]\( 1 \)[/tex] and [tex]\( -1 \)[/tex].
In conclusion, [tex]\( f(x) = \frac{x^2}{x^2 - 1} \)[/tex] is defined for all real numbers except [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], has a horizontal asymptote at [tex]\( y = 1 \)[/tex], vertical asymptotes at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex], and passes through the origin [tex]\( (0, 0) \)[/tex].
