# Thread: Finding range with limited domain

1. ## Finding range with limited domain

I am given the function y=2 times x^2 + 4
The domain is -3< x <5
I am to find the range.

My book (Sparknotes Math lvl 2 review book) says, "The best way to solve this type of problem is to manipulate the domain of x ine xactly the same way that x is manipulated in the function. First x is squared, then multiplied by 2, then aded to 4; we need to do the same thing to the bounds of the domain" Ok... so then they have this:

-3<x<5
0<x^2<25
0<2x^2< 50
4<2x^2 +4 < 54

I don't understand steps 2 and 3. -3 squared is 9. 9 times 2 is 18, plus 4 is 22. I want to sasy there is a typo, but when I graphed it, 4 appeared to be correct (unless I read the graph wrong). Is it a typo or am I missing something ?

2. ## Re: Finding range with limited domain

Originally Posted by benny92000
I am given the function y=2 times x^2 + 4
The domain is -3< x <5
I am to find the range.

My book (Sparknotes Math lvl 2 review book) says, "The best way to solve this type of problem is to manipulate the domain of x ine xactly the same way that x is manipulated in the function. First x is squared, then multiplied by 2, then aded to 4; we need to do the same thing to the bounds of the domain" Ok... so then they have this:

-3<x<5
0<x^2<25
0<2x^2< 50
4<2x^2 +4 < 54

I don't understand steps 2 and 3. -3 squared is 9. 9 times 2 is 18, plus 4 is 22. I want to sasy there is a typo, but when I graphed it, 4 appeared to be correct (unless I read the graph wrong). Is it a typo or am I missing something ?
This is a decreasing function so we need to look at it's minimum point to decide the range. For example $\displaystyle f(-1) = 6$ which is in the range.

f(x) has a minimum at x=0 (complete the square if you want to be sure) so the lowest point is (0,4) which is why the book has used x=0 in this example.

Personally, I'd have worked out the minimum point and used f(x) to find it's corresponding value in the range but the book's method does work

3. ## Re: Finding range with limited domain

-3<x<5
0<x^2<25
0<2x^2< 50
4<2x^2 +4 < 54
If this is from the book, then the book is mostly correct. Clearly, squaring does not map (-3, 5) into (18, 25) because the first interval contains 0. In fact, (-3, 5) = (-3, 0) ∪ [0, 5). On each of these two intervals, x^2 is monotone and 1-1, so (-3, 0) is mapped to (0, 9) and [0, 5) is mapped to [0, 25). Altogether, x^2 maps (-3, 5) to [0, 25). Note that 0 is included in the interval.

The rest is easy: doubling maps [0, 25) to [0, 50) and adding 4 changes it to [4, 54).

For me, it's easier to solve such problem looking at a graph.

4. ## Re: Finding range with limited domain

So you both would recommend that for this problem I find the minimum then graph to find maximum?

5. ## Re: Finding range with limited domain

I usually start by sketching a graph, but the book's method—"to manipulate the domain of x in exactly the same way that x is manipulated in the function"—can also work well. One must only keep in mind that a function f transforms [a, b] into [f(a), f(b)] (or similarly for open intervals) if f is continuous and monotonic on [a, b].

6. ## Re: Finding range with limited domain

Ok. But aren't we not exactly "manipulating the domain of x in exactly teh ame way that x is maniuplated in the function" if we are acknowledging that the minimum is 4? Do you have an example for the continuous functions?

7. ## Re: Finding range with limited domain

Originally Posted by benny92000
But aren't we not exactly "manipulating the domain of x in exactly teh ame way that x is maniuplated in the function" if we are acknowledging that the minimum is 4?
You may be right; why does it matter much? I am not claiming that there is another way which is essentially different and does not reduce to manipulating the domain. The important thing is that you can find the minimum and the range and prove that they are correct.

Another example let $\displaystyle f(x)= 1+\sin x$. Then $\displaystyle f(x)$ maps $\displaystyle [0,\pi/2)$ to $\displaystyle [1,2)$ because $\displaystyle f(x)$ is continuous and monotonic on the interval. On the other hand, it is wrong to say that $\displaystyle f(x)$ maps $\displaystyle [0,\pi]$ to $\displaystyle [f(0),f(\pi)]=[1,1]$ because $\displaystyle f(x)$ is not monotonic on $\displaystyle [0,\pi]$. This segment has to be divided into two intervals of monotonicity: $\displaystyle [0,\pi]=[0,\pi/2]\cup(\pi/2,\pi]$. The first interval is mapped to [1, 2] and the second to [1, 2), so the overall result is [1, 2].

Let $\displaystyle f(x) = 1/x$. Then [-1, 1] is not mapped into $\displaystyle [f(-1),f(1)]=[-1,1]$ because $\displaystyle f(x)$ is neither monotonic nor continuous on $\displaystyle [-1, 1]$ (it is not defined when $\displaystyle x = 0$). Again, we can divide $\displaystyle [-1, 1]$ into to intervals of monotonicity $\displaystyle [-1, 0)$ and $\displaystyle (0, 1]$, which are mapped to $\displaystyle (-\infty,-1]$ and $\displaystyle [1,\infty)$, respectively.