which statement describes the behavior of the function $f(x)=\\frac{2x}{1 - x^2}$?\n\n- the graph approaches…

which statement describes the behavior of the function $f(x)=\\frac{2x}{1 - x^2}$?\n\n- the graph approaches $-2$ as $x$ approaches infinity.\n- the graph approaches $0$ as $x$ approaches infinity.\n- the graph approaches $1$ as $x$ approaches infinity.\n- the graph approaches $2$ as $x$ approaches infinity.

which statement describes the behavior of the function $f(x)=\\frac{2x}{1 - x^2}$?\n\n- the graph approaches $-2$ as $x$ approaches infinity.\n- the graph approaches $0$ as $x$ approaches infinity.\n- the graph approaches $1$ as $x$ approaches infinity.\n- the graph approaches $2$ as $x$ approaches infinity.

Answer

Explanation:

Step1: Analyze the function for end behavior

To find the end behavior of ( f(x) = \frac{2x}{1 - x^2} ) as ( x \to \infty ), we can divide the numerator and denominator by the highest power of ( x ) in the denominator, which is ( x^2 ).

Step2: Divide numerator and denominator by ( x^2 )

[ \begin{align*} f(x) &= \frac{\frac{2x}{x^2}}{\frac{1}{x^2} - \frac{x^2}{x^2}}\ &= \frac{\frac{2}{x}}{\frac{1}{x^2} - 1} \end{align*} ]

Step3: Evaluate the limit as ( x \to \infty )

As ( x \to \infty ), ( \frac{2}{x} \to 0 ) and ( \frac{1}{x^2} \to 0 ). So, [ \lim_{x \to \infty} f(x) = \frac{0}{0 - 1} = 0 ]

Wait, there's a mistake here. Let's re - evaluate the division. The highest power of ( x ) in the denominator is ( x^2 ), but the numerator is ( 2x ) (degree 1) and the denominator is ( -x^2 + 1 ) (degree 2). For rational functions, when the degree of the numerator (( n )) is less than the degree of the denominator (( m )), the limit as ( x\to\pm\infty ) is 0. But let's do it again.

Wait, ( f(x)=\frac{2x}{1 - x^2}=\frac{2x}{-x^2 + 1} ). Divide numerator and denominator by ( x^2 ):

[ f(x)=\frac{\frac{2x}{x^2}}{\frac{-x^2}{x^2}+\frac{1}{x^2}}=\frac{\frac{2}{x}}{- 1+\frac{1}{x^2}} ]

As ( x\to\infty ), ( \frac{2}{x}\to0 ) and ( \frac{1}{x^2}\to0 ). So ( \lim_{x\to\infty}f(x)=\frac{0}{-1 + 0}=0 ). Wait, but let's check the options again. Wait, maybe I made a mistake in the degree analysis. Wait, the numerator is degree 1, denominator is degree 2. So the limit as ( x\to\infty ) should be 0. But let's check the options. One of the options is "The graph approaches 0 as x approaches infinity." But wait, let's re - express the function:

( f(x)=\frac{2x}{1 - x^2}=\frac{2x}{-(x^2 - 1)}=-\frac{2x}{x^2 - 1} )

As ( x\to\infty ), ( x^2-1\approx x^2 ), so ( f(x)\approx-\frac{2x}{x^2}=-\frac{2}{x}\to0 ). So the correct option is "The graph approaches 0 as x approaches infinity."

Wait, but let's check the options again. The options are:

  1. The graph approaches - 2 as x approaches infinity.

  2. The graph approaches 0 as x approaches infinity.

  3. The graph approaches 1 as x approaches infinity.

  4. The graph approaches 2 as x approaches infinity.

From our calculation, the limit is 0. So the correct option is the second one.

Wait, no, I think I messed up the division. Let's do it again. Let's factor out ( x^2 ) from the denominator:

( f(x)=\frac{2x}{1 - x^2}=\frac{2x}{-x^2(1-\frac{1}{x^2})}=\frac{\frac{2x}{-x^2}}{1 - \frac{1}{x^2}}=\frac{-\frac{2}{x}}{1-\frac{1}{x^2}} )

As ( x\to\infty ), ( -\frac{2}{x}\to0 ) and ( \frac{1}{x^2}\to0 ). So ( \lim_{x\to\infty}f(x)=\frac{0}{1 - 0}=0 ). So the correct option is "The graph approaches 0 as x approaches infinity."

Answer:

The graph approaches 0 as x approaches infinity.