How to find the range of #x^2/(1-x^2)#?

Question

I thought that the answer is #R:(-oo,-1)uu(-1,+oo)#but the book says that the correct answer is #R:(-oo,-1)uu[0,+oo)#
Can anyone help please?

Cora Alexander · Accepted Answer

The range of #x^2/(1-x^2)# is #(-oo, -1) uu [0, oo)#

Let:

#y = x^2/(1-x^2)#

and solve for #x#...

Multiplying both sides by #1-x^2#, we get:

#y-yx^2 = x^2#

Adding #yx^2# to both sides, this becomes:

#y = (y+1)x^2#

Then dividing both sides by #(y+1)# we get:

#x^2 = y/(y+1)#

This has solutions if and only if:

#y/(y+1) >= 0#

That is, if either of the following:

#y >= 0" "# and #" "y + 1 > 0#. That is #y >= 0#

#y <= 0" "# and #" "y + 1 < 0#. That is #y < -1#

So the range of #x^2/(1-x^2)# is #(-oo, -1) uu [0, oo)#

graph{x^2/(1-x^2) [-10, 10, -5, 5]}

Christian Antunez · Answer

undefined To find the range of the function $ f(x) = \frac{x^2}{1 - x^2} $, we need to consider the restrictions on the domain that ensure the function is well-defined.

The denominator $ 1 - x^2 $ cannot be equal to zero because division by zero is undefined. Therefore, we must find the values of $ x $ that make $ 1 - x^2 $ equal to zero.

Solving $ 1 - x^2 = 0 $ gives us $ x^2 = 1 $.

This equation has two solutions: $ x = 1 $ and $ x = -1 $.

Thus, the function is undefined when $ x = 1 $ and $ x = -1 $.

Next, let's consider the behavior of the function as $ x $ approaches these values.

As $ x $ approaches $ 1 $ or $ -1 $ from the left, $ 1 - x^2 $ becomes positive, and as $ x $ approaches $ 1 $ or $ -1 $ from the right, $ 1 - x^2 $ becomes negative. This means that $ f(x) $ approaches positive infinity as $ x $ approaches $ -1 $ from the left, and negative infinity as $ x $ approaches $ -1 $ from the right. Similarly, $ f(x) $ approaches positive infinity as $ x $ approaches $ 1 $ from the right, and negative infinity as $ x $ approaches $ 1 $ from the left.

Therefore, the range of the function $ f(x) = \frac{x^2}{1 - x^2} $ is all real numbers except $ -\infty $ to $ -1 $, $ 1 $, and $ 1 $ to $ \infty $. In interval notation, the range is $ (-\infty, -1) \cup (-\infty, 1) \cup (1, \infty) $.

How to find the range of #x^2/(1-x^2)#?

I thought that the answer is #R:(-oo,-1)uu(-1,+oo)#but the book says that the correct answer is #R:(-oo,-1)uu[0,+oo)# Can anyone help please?

I thought that the answer is #R:(-oo,-1)uu(-1,+oo)#but the book says that the correct answer is #R:(-oo,-1)uu[0,+oo)#
Can anyone help please?