Download MA 107 Fall 2012 Jessica Wagner Domain/Range Problems

MA 107 Fall 2012
Jessica Wagner
Domain/Range Problems - August 25, 2011
What are the domain and range of the following functions? (Detailed explanations are found in the solution key.)
√
1. f (x) = x + 6
Solution: We have a square root here, so we need to make sure that any value of x that we put in there will not
let the square root be undefined. Thus, x + 6 ≥ 0 → x ≥ −6, and our domain is [−6, ∞) .
Square roots are also a heads up that our range for that part of the function will be greater or equal to 0, and since
we don’t have any other parts of the function, our range is [0, ∞) .
2. f (x) = 51 x2 + 6
Solution: When we approach these functions, it is best to start thinking as if our domain and range are all real
numbers. Here we have no square roots or division, so we could plug any value of x in that we wanted. There
are no x issues. Thus our domain is all real numbers, or (−∞, ∞) .
Range is a little trickier. We need to figure out what numbers will come out when we plug numbers in for
x. We focus first on the fact that the only x we have in the function is x2 . We know that any number squared
must be greater than or equal to 0. So we can think about this mathematically by claiming that f (x) ≥ 51 ∗ 0 + 6,
because this is the lowest possible value for the x part of the function. Thus our value for f (x) cannot get lower
than 6, and our range is [6, ∞) .
3. f (x) =
3x2 −2
x−3
Solution: Here we have a fraction, so we need to check the denominator for any values of x which make it
equal 0. Note that this doesn’t mean that the denominator can’t be negative, it just can’t be 0. Thus we set up:
S
x − 3 6= 3 → x 6= 3. Thus our domain is (−∞, 3) (3, ∞), or all real numbers except 3 .
Now range. We note that there is an x2 on the top part of the fraction, which may lead us to believe that
only positive numbers would be our f(x). However, the x on the bottom will cause us to still have numbers from
both positive and negative sides of the number line. For example, if we plug in 5 for x, we get 3∗25−2
→ 73
5−3
2 ,
3∗25−2
−73
a positive number. If we plug in -5 for x, we get −5−3 → 8 , a negative number. Increased values in either
direction of x would increase or decrease our f(x). We also should
q check that it’s possible for f(x) to equal 0. To
do so, we can set f (x) = 3x2 − 2 = 0 → 3x2 = 2 → x = 23 . Even if you don’t carry the math through this
far, if you do so long enough to see that x can actually equal something to make it true, you can claim that 0 is
included in the range. Basically this is going to happen anytime you have x’s appearing in the top of a fraction.
Thus, our range here is all real numbers .
4. f (x) =
p
x(x − 2)
Solution: Note here that the stuff under the square root has been factored to kind of give you
√ a hint as to how
to figure out the domain here. We would have the same procedure for the equation f (x) = x2 − 2x, we just
would have to factor that to get it to this point. We still have a square root here, and thus its argument must be
greater than or equal to 0. So x(x − 2) ≥ 0. In case you haven’t seen this before, when we have an equation
like this, we can take each piece and individually set it equal to 0 (actually greater than or equal in our cases, but
it works the same). So we first take x ≥ 0. That one is easy. Then we see that x − 2 ≥ 0 → x ≥ 2. While it
would be tempting to say that our domain would be [0.∞), we actually can only use the numbers that appear in
both intervals here, and they only overlap from 2 to infinity. So our domain is [2, ∞) .
Luckily, range is not too bad here. Square root means not negative, so our range is [0, ∞) .
MA 107 Fall 2012
Jessica Wagner
Domain/Range Problems - August 25, 2011
5. f (x) = 3x + 2
Solution: Do we have any square roots or divisions by x? No. So we can put in any x value we’d like and a
viable answer would come out. Our domain is all real numbers .
Do we have any squared x’s or square roots? Nope. It would be possible to get any value of f(x) from the
right x, so our range is all real numbers here as well.
For these functions, the solutions key has less detailed explanations, but still check yourself to make sure you’re
on the right track.
6. f (x) =
√
1
x2 −x−6
Solution: Both square root and denominator issues.
x2 − x − 6 > 0 → (x − 3)(x + 2) > 0 → x − 3 > 0 & x + 2 > 0. So x > 3 & x > −2. These overlap in the
domain x > 3, or (3, ∞) . Note the parenthesis instead of bracket.
Bottom of the fraction will always be positive. Top of the fraction will too because it contains no x’s. We
also need to note that since there are no x’s in the top of the fraction, f(x) can never be 0. Thus the range is
(0, ∞) .
7. f (x) =
√
x−4+2
Solution: Square root an issue. x − 4 ≥ 0 → x ≥ 4. So domain is x ≥ 4 or [4, ∞) .
The square root part of the function will never go below 0. So f (x) ≥ 0 + 2 ≥ 2. Our range is [2, ∞) .
8. f (x) =
x
x−3
Solution: Domain is x 6= 3 .
Range is all real numbers .
9. f (x) =
√
3x − 6
Solution: Domain is x ≥ 2 .
Range is [0, ∞) .
10. f (x) =
√3
x−2
Solution: Domain is (2, ∞) .
Range is all real numbers except 0 .
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MA 107 Fall 2012
Jessica Wagner
√
11. f (x) = − x + 10
Solution: Domain is [−10, ∞) .
Range is (−∞, 0] .
√
12. f (x) = − 4x − 12 + 7
Solution: Domain is [3, ∞) .
Range is (−∞, 7] .
13. f (x) =
2x−4
x−2
+3
Solution: Domain is x 6= 2 .
Range is all real numbers .
14. f (x) = x2 − 15
Solution: Domain is all real numbers .
Range is [−15, ∞) .
15. f (x) = 4x2 + 5
Solution: Domain is all real numbers .
Range is [5, ∞) .
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Domain/Range Problems - August 25, 2011