How do you solve this precalculus problem with domains?

Question

Find the domain of g(x)= square root of (x^3-9x) ??

Can you also explain the steps if you can? None of the youtube videos are helping me. All of them are doing problems with squares and not cubes.

Answer 1

g(x) = √(x³-9x)            ← Square roots are real only if the radicand is non-negative.

                                     So, the domain will be all values of x that make x³-9x ≥ 0



So, we're looking for the solution to

      x³ - 9x ≥ 0

    x(x² - 9) ≥ 0

x(x+3)(x-3) ≥ 0        ← The zeros of x(x+3)(x-3) are -3, 0, 3.

                                  Use these three numbers to divide the number line into four intervals.



                  Test

Intervals    Number         x(x+3)(x-3) > 0       T/F

——————————————————————

(-inf,-3)        -100       neg(neg)(neg) > 0     False

(  -3 , 0)            -1       neg(pos)(neg) > 0     True     ✔

 (  0 , 3)              1       pos(pos)(neg) > 0     False

(  3 , inf)         100        pos(pos)(pos) > 0     True     ✔



                                AND



                         x(x+3)(x-3) = 0

                           x = 0 , -3 , 3     ✔



                                   ⇓



                              ANSWER

                           [-3,0] ⋃ [3,inf)



Have a good one!

.

Answer 2

The domain is all real numbers, unless there is some operation which is not legal for all x.



In this case it's the square root. You can only take a square root of something which is >= 0. So the domain is the set of x such that x^3 - 9x >= 0.



Now you have to solve that inequality to find out what x that is.



x^3 - 9x >= 0

x(x^2 - 9) >= 0

x(x - 3)(x + 3) >= 0

There are three factors here. The place where each one is 0 defines a point where the factor changes from negative to positive. For instance the factor x is negative when x < 0 and positive when x > 0.

The factor x + 3 is negative when x < -3 and positive when x > -3.

The factor x - 3 is negative when x < 3 and positive when x > 3.



So you can see those three factors define three "critical points" x = -3, 0, 3 where the signs change. They divide the real number line up into separate regions where the factors have different combinations of signs.



You want them to multiply to be a non-negative number. So the number of negative factors has to be either 0 or 2. If you have 1 negative number or 3 negative numbers, the product is negative.



Put all that together. Find the places defined by x = -3, 0, 3 where there are either 0 or 2 negative factors.

Answer 3

g(x)= square root of (x^3-9x)



answers for sq root cannot be negative



x^3 - 9x = 0

x(x+3)(x-3) = 0

x = -3, 0, and +3

test each region , (-infinity,-3] and [-3,0] and [0,+3] and [+3, +infinity)

for x = -10 ... x^3 - 9x = -1000 + 90 = -910 <<< not in domain

for x = -1 ... -1^3 -9(-1) = +8 <<< yes in domain

for x = 1 ... 1^3 -9(1) = -8 <<< not in domain

for x = 10 ... 10^3 -9(10) = 910 << yes in domain



domain .. [-3,0] and [+3, +infinity)

Answer 4

(x^3-9x) > or = 0

---> x in [-3,0]U[3,infinity)



(by the graph of y=x^3-9x)