1. ## invalid range

Is it possible to have a range of $\left[\frac{1}{4}, \infty\right]$ ?

I mean... you can't have something equal to infinity right, because there's always going to be something bigger... but I'm wondering if they did it like this because weird stuff happens at asymptotes.
This is the question that has it. I know how to do the question, but I don't know about my above question.

A function with rule $f(x)=x^{-2}$ can be defined on different domains. In the list below, the first set represents the domain selected and the second set represents the corresponding range, but in one case the range is incorrect. Which pair does not have the correct range for the given domain?

A. Domain: $[-2,-1]$, Range: $\left[\frac{1}{4},1\right]$

B. Domain: $[-2,0) \cup (0,1]$, Range: $\left[\frac{1}{4}, \infty\right]$

C. Domain: $[-1,1] \setminus \lbrace{0\rbrace}$, Range: $[1,\infty)$

D. Domain: $[-1,2] \setminus \lbrace{0\rbrace}$, Range: $\left[\frac{1}{4},1\right]$

E. Domain: $[1,2)$, Range: $\left(\frac{1}{4},1\right]$

[1998 MM]

Thanks

2. Originally Posted by DivideBy0
Is it possible to have a range of $\left[\frac{1}{4}, \infty\right]$ ?

I mean... you can't have something equal to infinity right, because there's always going to be something bigger... but I'm wondering if they did it like this because weird stuff happens at asymptotes.
This is the question that has it. I know how to do the question, but I don't know about my above question.

A function with rule $f(x)=x^{-2}$ can be defined on different domains. In the list below, the first set represents the domain selected and the second set represents the corresponding range, but in one case the range is incorrect. Which pair does not have the correct range for the given domain?

A. Domain: $[-2,-1]$, Range: $\left[\frac{1}{4},1\right]$

B. Domain: $[-2,0) \cup (0,1]$, Range: $\left[\frac{1}{4}, \infty\right]$

C. Domain: $[-1,1] \setminus \lbrace{0\rbrace}$, Range: $[1,\infty)$

D. Domain: $[-1,2] \setminus \lbrace{0\rbrace}$, Range: $\left[\frac{1}{4},1\right]$

E. Domain: $[1,2)$, Range: $\left(\frac{1}{4},1\right]$

[1998 MM]

Thanks
Hello,

compare the answers B. and C. In both cases the domains contain values left and right of zero. Then the upper limits of the ranges must be equal. So if B. is correct then C. is wrong or if C. is correct then B. is wrong.
Since a range contains values and the infinity is not a number B. is wrong.

3. Yes but if you said that in the exam you would get it wrong.

According to the paper, D is the correct answer. So this implies that B was a typo? Good thing I'm not doing the 1998 exam.

4. I agree that D is incorrect, for example if your domain goes from -1 to 2, then .5 is in your domain, and 1/.5^2=1/.25=4 But 4 is outside the given range of 1/4 to 1

And I think when they say the range goes to infinity, that means there is no upper limit to what the y value can be. Because if you keep getting closer and closer to zero, your answer will keep getting higher and higher.

Lets plug some x values into the equation $\frac{1}{x^2}$
.1 => 100
.01 => 10,000
.001 => 1,000,000
.0001 => 100,000,000
.00001 => 10,000,000,000
.000001 => 1,000,000,000,000

As you see the value is growing without bound, so it is approaching infinity. So it's range must have an upper limit of infinity (and since you cannot reach infinity, it must have a parentheses instead of a bracket)

5. Here is a graph showing what the function looks like. Notice that as it gets closer to zero, the range shoots up towards infinity. So any answer who's domain is greater than or less than zero (but never equal to zero, that is not in the domain of the function... well... with a username like yours, maybe it is ) will have to have infinity as an upper limit. And when I say greater than or less than, I mean one boundary is zero. such as (0, 5] or (-1, 0) or [-20,0)U(0,3) these are all domains which will have infinity as their upper limit for the range (note, I made them red to emphasize what we are looking at, it's purely aesthetic, no technical meaning to it).

edit: to emphasize why D is incorrect, try considering the domain of f(x)=x^2 from [-5, 5]. If you simply allow the left and right domain values to be the boundaries of your range, it appears that your range is [25,25] which means it must be a straight line of y=25. But this is clearly not the case, because 2 is in the domain, and 2's y value is 4, which is outside the range of [25,25], so we can't just assess the y values at the boundaries of the domain as they did in example D, we must account for every y value in between.