Let's find the range of the function [tex]\( h(x) = \frac{x^2}{1 - x^2} \)[/tex].
To understand the behavior of this function, we need to consider where it is defined and its behavior in different intervals of [tex]\( x \)[/tex]. Specifically, the function [tex]\( h(x) \)[/tex] becomes undefined when the denominator [tex]\( 1 - x^2 \)[/tex] equals zero.
### Step-by-Step Solution:
1. Determine where the function is undefined:
[tex]\[
1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1
\][/tex]
Hence, the function is undefined at [tex]\( x = 1 \)[/tex] and [tex]\( x = -1 \)[/tex].
2. Analyze intervals excluding [tex]\(\pm 1\)[/tex]:
- Interval 1: [tex]\( x \in (-\infty, -1) \)[/tex]
For [tex]\( x \)[/tex] in this interval, [tex]\( x^2 > 1 \)[/tex], and [tex]\( 1 - x^2 < 0 \)[/tex]. Thus, the function [tex]\( h(x) \)[/tex] will produce positive values but since both [tex]\( x^2 \)[/tex] and [tex]\( (1-x^2) \)[/tex] have opposite signs, the result of [tex]\( h(x) \)[/tex] will approach infinity as [tex]\( x^2 \)[/tex] becomes very large.
- Interval 2: [tex]\( x \in (-1, 1) \)[/tex]
For [tex]\( x \)[/tex] in this interval, [tex]\( x^2 < 1 \)[/tex], and [tex]\( 1 - x^2 > 0 \)[/tex]. Here, the function [tex]\( h(x) \)[/tex] produces positive values ranging from 0 up to, but not including, infinity as [tex]\( x \)[/tex] approaches 1 or -1 from the inside of this interval.
- Interval 3: [tex]\( x \in (1, \infty) \)[/tex]
For [tex]\( x \)[/tex] in this interval, [tex]\( x^2 > 1 \)[/tex], and [tex]\( 1 - x^2 < 0 \)[/tex]. Similarly to Interval 1, [tex]\( h(x) \)[/tex] will produce positive values, and as [tex]\( x \)[/tex] grows larger, the values of [tex]\( h(x) \)[/tex] will again approach infinity.
3. Conclusion on the range:
This detailed interval analysis shows that the function produces:
- Negative values when approaching [tex]\( -\infty \)[/tex]
- Positive values approaching 0 and then positive infinity in the interval [tex]\( (-1, 1) \)[/tex]
- Values approaching infinity for [tex]\( x\)[/tex] beyond 1.
From this analysis, we can conclude that the function covers all real numbers except at [tex]\( -1 \)[/tex] and [tex]\( 1 \)[/tex].
Thus, the range of [tex]\( h(x) \)[/tex] is given by:
[tex]\[
(-\infty, -1) \cup (-1, 1) \cup (1, +\infty)
\][/tex]
### The best answer from the given choices is:
b. [tex]$(-\infty,-1) \cup(-1,1) \cup(1,+\infty)$[/tex]
