f(x)=1/x^2 on-3<=x<=3. What is the range of f(x) on this interval?

Question

I need help on my Algebra II homework.

Answer 1

The range is the set of possibilities for what f(x) can be. [The domain, which you have been given (-3 ≤ x ≤ 3), is the set of possibilities for what x can be.]

 

Are there any gaps in the graph of this function? One thing that may cause a gap to exist is a place where a certain value of x would cause a division by zero, which is undefined. [The function is not defined, for that value of x.] Here, if x = 0, then 1 / x² = "1 / 0", which is undefined, so we can ignore x = 0.

 

If you have a graphing calculator, you can use it to draw the graph of f(x). Try looking at the values of f(x) for x = -3 and 3 (the bounds of your given domain). For both of these values, f(x) = 1 / 9. What happens for other values, and what happens as x approaches 0, from either side? You won't be able to input x = 0, but you'll be able to see what happens, as you get closer and closer to it. The value of f(x) increases without bound, so it goes to infinity (∞).

 

Thus, the range of f(x) is 1 / 9 ≤ f(x) < ∞. Why not ≤ ∞? You can't actually get the infinity; you can only approach it, as x approaches 0.