# uc3rmw_0103e05

0 (0 Likes / 0 Dislikes)

Hi there.
In this example, we will be finding
the domain of the following functions.
In part A we have f of x equal
to 1 over 1 minus x squared.
In B we have g of x equal to
the square root of x plus 1.
In part C, h of x equal 1 over
the square root of x minus 1.
In and part D, p of x equal to the
square root of x squared minus x minus 6.
The domain of the function
is defined as the real number
values for where the function is defined.
The first function, given in part A, is
f of x equal 1 over 1 minus s squared.
This is a rational function.
And for rational functions,
the function is defined
for values of x for which the
denominator is not equal to 0.
Therefore, the domain for this
function will be all real numbers
x excluding those for which the denominator,
which is 1 minus x squared, is equal to 0.
To determine those values that makes
the denominator 1 minus x squared
equal to 0 we solve the equation 1
minus x squared equals 0 as follows.
1 minus x squared equals 0.
We factor the left-hand side of the equation
using the difference of squares formula.

So this is 1 plus x times 1 minus x equal to 0. We're now going to set each factor equal to 0 and solve each linear equation. First linear equation is 1 plus x equals 0. Then we have 1 minus x equal to 0 for the other linear equation. So the solutions are x equal to minus 1, or x equal to 1. So the denominator 1 minus x squared is equal to 0 when x equals plus or minus 1. Therefore, the domain of f in set notation is all values of x such that x is not equal to plus or minus 1. We can also write that in interval notation as minus infinity to minus 1 union minus 1 to plus 1 union 1 to infinity. The second function, given in part B, is g of x equals the square root of x plus 1. This is a square root function. And for square root functions, the function is defined for values of x for which the argument of the square root is not negative. For this square root function the argument of the square root is x plus 1. x plus 1 cannot be negative, because we cannot take the square root of a negative number. Therefore, the domain for this function will be all real numbers for x for which the argument of the square root-- x plus 1-- is non-negative or greater than or equal to 0. To determine the values that make the argument of the square root greater than or equal to 0, we solve the inequality condition x plus 1 greater than or equal to 0. Subtracting 1 from both sides will solve the inequality. We have x is greater than or equal to minus 1. Thus, the domain of g is the set of numbers x all values of x such that x is greater than or equal to minus 1. In interval notation we have minus 1 to infinity, where there's a bracket on the left part of the interval to indicate that minus 1 is included in the domain for this function. The third function, given in part C, is h of x equal to 1 over the square root of x minus 1. This is a rational function, as well, just like we had in part A. But the denominator is a square root like we worked out in part B. Thus, both approaches we carried out in parts A and B for finding the domain apply for this function. The domain of this function will exclude those values of x for which the denominator is 0, as well as constrain the domain to values of x for which the argument of the square root is not negative. Because of the square root, the domain is constrained to real numbers values of x in which the argument of the square root x minus 1 is greater than or equal to 0. So because of the square root, we have x minus 1 is greater than or equal to 0. And we can solve this inequality. And we get that x is constrained to be values of x greater than or equal to 1. So because the denominator cannot equal 0, so x minus 1 cannot equal 0. Adding 1 to both sides, we have that x cannot equal 1. Combining both conditions, we have that the domain is constrained to values of x where x minus 1 is greater than 0. Adding 1 to both sides to solve the inequality, we have that the domain is x greater than 1. In set notation, we can express this as all real values x such that x is greater than 1. Or in interval notation, we can express that as the interval from 1 to infinity where both endpoints of that interval get parentheses indicating that the 1 is not included in the domain. The fourth function, given in part D, is p of x equals the square root of x squared minus x minus 6. Like the function g in part B, this is a square root function as well. Again, for square root functions, the function is defined for values of x for which the argument of the square root is not negative. For this square root function the argument of the square root is x squared minus x minus 6. This cannot be negative, because we cannot take the square root of a negative number. Therefore, the domain of this function will be all real numbers for x for which the argument of the square root-- again x squared minus x minus 6-- is non-negative or greater than or equal to 0. To determine the values that make the argument of the square root greater than or equal to 0, we solve the inequality x squared minus x minus 6 greater than or equal to 0. First we factor the expression on the left-hand side of the inequality as x plus 2 times x minus 3 greater than or equal to 0. Now we apply the test point method. And the boundary points that define the intervals we test in the test method are obtained from solving the equation x plus 2 times x minus 3 equal to 0. Setting each linear factor equal to 0, we get x equal to minus 2, or x equal to 3. These are the interval boundaries that define the test intervals that are minus infinity to minus 2; minus 2 to 3; and finally, 3 to infinity. On each of these intervals we choose a test point and evaluate x plus 2, x minus 3 to determine its sign on that interval. We arbitrarily choose the test point x equals minus 3 on the interval from negative infinity to minus 2, x equals 0 on the interval from minus 2 to 3, and x equal 4 on the interval from 3 to infinity. For the test point x equal minus 3 on the interval from negative infinity to minus 2, we have x equal minus 3 is our test point again on minus infinity to minus 2 that interval. So let's evaluate the first test point. We're going to plug x equal minus 3 into x plus 2, x minus 3, and evaluate that expression. We have minus 3 plus 2 times minus 3 minus 3, which simplifies to minus 1 times minus 6 or 6. Now 6 is a positive number. Hence, x plus 2, x minus 3 is positive over the interval from minus infinity to minus 2. Next for the test point x equal to 0 on the interval from minus 2 to 3, we evaluate x plus 2, x minus 3 at that test point. We have 0 plus 2 times 0 minus 3. Simplifying, that gives us 2 times minus 3, or minus 6, which is a negative number. And x plus 2, x minus 3 is negative over the interval from minus 2 to 3. And for the last test point x equal 4 on the interval from 3 to infinity, we have x plus 2, x minus 3 evaluated at x equals 4 is 4 plus 2 times 4 minus 3, which simplifies to 6 times 1, or 6, which is positive. Hence, x plus 2, x minus 3 is positive over the interval from 3 to infinity. So the figure again shows the intervals on the real number line. And the sine of x plus 2, x minus 3 over the three intervals. The expression x plus 2, x minus 3 is greater than or equal to 0 only on the intervals from minus infinity to minus 2 and the interval from 3 to infinity. Hence the domain of p in the interval notation is minus infinity to minus 2 union the interval from 3 to infinity.

So this is 1 plus x times 1 minus x equal to 0. We're now going to set each factor equal to 0 and solve each linear equation. First linear equation is 1 plus x equals 0. Then we have 1 minus x equal to 0 for the other linear equation. So the solutions are x equal to minus 1, or x equal to 1. So the denominator 1 minus x squared is equal to 0 when x equals plus or minus 1. Therefore, the domain of f in set notation is all values of x such that x is not equal to plus or minus 1. We can also write that in interval notation as minus infinity to minus 1 union minus 1 to plus 1 union 1 to infinity. The second function, given in part B, is g of x equals the square root of x plus 1. This is a square root function. And for square root functions, the function is defined for values of x for which the argument of the square root is not negative. For this square root function the argument of the square root is x plus 1. x plus 1 cannot be negative, because we cannot take the square root of a negative number. Therefore, the domain for this function will be all real numbers for x for which the argument of the square root-- x plus 1-- is non-negative or greater than or equal to 0. To determine the values that make the argument of the square root greater than or equal to 0, we solve the inequality condition x plus 1 greater than or equal to 0. Subtracting 1 from both sides will solve the inequality. We have x is greater than or equal to minus 1. Thus, the domain of g is the set of numbers x all values of x such that x is greater than or equal to minus 1. In interval notation we have minus 1 to infinity, where there's a bracket on the left part of the interval to indicate that minus 1 is included in the domain for this function. The third function, given in part C, is h of x equal to 1 over the square root of x minus 1. This is a rational function, as well, just like we had in part A. But the denominator is a square root like we worked out in part B. Thus, both approaches we carried out in parts A and B for finding the domain apply for this function. The domain of this function will exclude those values of x for which the denominator is 0, as well as constrain the domain to values of x for which the argument of the square root is not negative. Because of the square root, the domain is constrained to real numbers values of x in which the argument of the square root x minus 1 is greater than or equal to 0. So because of the square root, we have x minus 1 is greater than or equal to 0. And we can solve this inequality. And we get that x is constrained to be values of x greater than or equal to 1. So because the denominator cannot equal 0, so x minus 1 cannot equal 0. Adding 1 to both sides, we have that x cannot equal 1. Combining both conditions, we have that the domain is constrained to values of x where x minus 1 is greater than 0. Adding 1 to both sides to solve the inequality, we have that the domain is x greater than 1. In set notation, we can express this as all real values x such that x is greater than 1. Or in interval notation, we can express that as the interval from 1 to infinity where both endpoints of that interval get parentheses indicating that the 1 is not included in the domain. The fourth function, given in part D, is p of x equals the square root of x squared minus x minus 6. Like the function g in part B, this is a square root function as well. Again, for square root functions, the function is defined for values of x for which the argument of the square root is not negative. For this square root function the argument of the square root is x squared minus x minus 6. This cannot be negative, because we cannot take the square root of a negative number. Therefore, the domain of this function will be all real numbers for x for which the argument of the square root-- again x squared minus x minus 6-- is non-negative or greater than or equal to 0. To determine the values that make the argument of the square root greater than or equal to 0, we solve the inequality x squared minus x minus 6 greater than or equal to 0. First we factor the expression on the left-hand side of the inequality as x plus 2 times x minus 3 greater than or equal to 0. Now we apply the test point method. And the boundary points that define the intervals we test in the test method are obtained from solving the equation x plus 2 times x minus 3 equal to 0. Setting each linear factor equal to 0, we get x equal to minus 2, or x equal to 3. These are the interval boundaries that define the test intervals that are minus infinity to minus 2; minus 2 to 3; and finally, 3 to infinity. On each of these intervals we choose a test point and evaluate x plus 2, x minus 3 to determine its sign on that interval. We arbitrarily choose the test point x equals minus 3 on the interval from negative infinity to minus 2, x equals 0 on the interval from minus 2 to 3, and x equal 4 on the interval from 3 to infinity. For the test point x equal minus 3 on the interval from negative infinity to minus 2, we have x equal minus 3 is our test point again on minus infinity to minus 2 that interval. So let's evaluate the first test point. We're going to plug x equal minus 3 into x plus 2, x minus 3, and evaluate that expression. We have minus 3 plus 2 times minus 3 minus 3, which simplifies to minus 1 times minus 6 or 6. Now 6 is a positive number. Hence, x plus 2, x minus 3 is positive over the interval from minus infinity to minus 2. Next for the test point x equal to 0 on the interval from minus 2 to 3, we evaluate x plus 2, x minus 3 at that test point. We have 0 plus 2 times 0 minus 3. Simplifying, that gives us 2 times minus 3, or minus 6, which is a negative number. And x plus 2, x minus 3 is negative over the interval from minus 2 to 3. And for the last test point x equal 4 on the interval from 3 to infinity, we have x plus 2, x minus 3 evaluated at x equals 4 is 4 plus 2 times 4 minus 3, which simplifies to 6 times 1, or 6, which is positive. Hence, x plus 2, x minus 3 is positive over the interval from 3 to infinity. So the figure again shows the intervals on the real number line. And the sine of x plus 2, x minus 3 over the three intervals. The expression x plus 2, x minus 3 is greater than or equal to 0 only on the intervals from minus infinity to minus 2 and the interval from 3 to infinity. Hence the domain of p in the interval notation is minus infinity to minus 2 union the interval from 3 to infinity.