SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says. <pre> Find the domain of the function. ___________ _ y = &#8730;2x² + 5x - 12

Algebra ->  Radicals -> SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says. <pre> Find the domain of the function. ___________ _ y = &#8730;2x² + 5x - 12       Log On


   

Question 597391: I am stuck on this problem. My answer is different from what the textbook answer says. 
Find the domain of the function.

     _____________
y = √2x² + 5x - 12

Found 2 solutions by rapaljer, AnlytcPhil:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Values of x must be such that the quantities inside the radical must be GREATER THAN OR EQUAL TO 0. You have correctly identified the endpoints of the intervals, which are x=-4 and x=3/2. In my opinion, the easiest way to solve this domain problem is to use a GRAPHING CALCULATOR to graph y=sqrt%282x%5E2%2B5x-12%29. Then, from the graph, determine the values of x for which you have points on the graph. Here is the graph:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C+sqrt%282x%5E2%2B5x-12%29%29

Notice that the values of x that generate points on this graph are either to the LEFT of x=-4 or to the RIGHT of x=1.5 or 3/2.

Therefore the domain is all values of x that are less than or equal to -4, OR values of x that are greater than or equal to 3/2. In interval notation, this can be written (-inf,-4]U[3/2, inf).

The alternative to using a graphing calculator is to solve the quadratic inequality 2x%5E2%2B5x-12%3E=0. Most people have trouble trying to solve quadratic inequalities, so don't feel bad! I do NOT recommend this method. I think the graphing calculator is MUCH easier!!!

For additional explanation on FUNCTIONS, DOMAIN, and RANGE, please see my own website. The easiest way to find it is to use the easy to spell and easy to remember link www.mathinlivingcolor.com. On this single page website, there is a link at the bottom of the page that takes you to my Homepage.

Near the top of my homepage, look for the link "Basic, Intermediate, and College Algebra: One Step at a Time", and choose College Algebra (Chapter 2). Look for Section 2.07 Functions, Domain, and Range. Be sure to see the "Math in Living Color" explanations that go with this section, since it includes solutions using the graphing calculator! Actually, you may want to START out by looking at the Math in Living Color page for this section. I think you will like it.

Anyway, you will find explanations that may be easier to understand than your traditional textbooks. There are also VIDEOS of me teaching this topic a few years before I retired. These videos are FREE, and you can see them by looking on my Homepage for the link "Rapalje Videos in Living Color." There are videos on Functions, Domain, and Range in College Algebra (or easier problems on the Intermediate Algebra video).

If you like my website, please tell your friends and family. It's all FREE. You can contact me by Email at rapaljer@seminolestate.edu.

Dr. Robert J. Rapalje, Retired
Professor of Mathematics
Seminole State College of Florida





Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!

I think the following is more in line with what your teacher has in mind:
     _____________
y = √2x² + 5x - 12

2x² + 5x - 12 ≧ 0

(x + 4)(2x - 3) ≧ 0

Now we find critical numbers, which are the 
zeros of the left side, by setting each factor
equal to 0:

x + 4 = 0;  2x - 3 = 0
    x = -4;      x = 3/2 = 1 1/2

We place these critical number on a number line: 

<========⚫--------------------⚫--------------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5 

We use darkened circles instead of open circles because the 
critical numbers are themselves solutions in this case since 
the inequality is ≧ rather that >.

There are three intervals to test for solutions to the
inequality, the intervals between and beyond the critical
numbers

(-oo,-4), (-4, 3/2), and (3/2,oo).  Or if you haven't had
interval notation, then they are these inequalities:

x ≦ -4,  -4 ≦ x ≦ 3/2,  x ≧ 3/2

Choose a test point in the interval (-oo,-4), say -5.
Substitute 5 into the inequality:

    (x + 4)(2x - 3) ≧ 0
(-5 + 4)(2(-5) - 3) ≧ 0 
      (-1)(-10 - 3) ≧ 0
          (-1)(-13) ≧ 0
                 13 ≧ 0

That is true, so (-oo,-4) is part of the solution, so we
shade that interval on the number line:

 Choose a test point in the interval (-4,3/2), say 0.
Substitute 0 into the inequality:

    (x + 4)(2x - 3) ≧ 0
  (0 + 4)(2(0) - 3) ≧ 0 
         (4)(0 - 3) ≧ 0
            (4)(-3) ≧ 0
                -12 ≧ 0

That is false, so (-4,3/2) is NOT part of the solution, so we
DO NOT shade that interval on the number line, so we still have 
jus:t

<========⚫--------------------⚫--------------
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5 

Choose a test point in the interval (3/2,-oo), say 2.
Substitute 2 into the inequality:

    (x + 4)(2x - 3) ≧ 0
  (2 + 4)(2(2) - 3) ≧ 0 
         (6)(4 - 3) ≧ 0
             (6)(1) ≧ 0
                  6 ≧ 0

That is true, so (3/2,oo) is part of the solution, so we
shade that interval on the number line:

<========⚫--------------------⚫==============>
-6  -5  -4  -3  -2  -1   0   1   2   3   4   5


So the domain written in interval notation is

(-oo,-4] U [3/2, oo)

or as an inequality the domain is written this way:

x ≦ -4 OR x ≧ 3/2

Edwin