SOLUTION: Good Evening! I am working on some homework and ran into a problem that I am having trouble figuring out. This function stuff really has me confused.
Find the domain of the
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Question 377909: Good Evening! I am working on some homework and ran into a problem that I am having trouble figuring out. This function stuff really has me confused.
Find the domain of the function: h*(x)= 7x/x*(x^2-9)
Thank you so much!!
Denise Bissell
Found 3 solutions by stanbon, nyc_function, edjones:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the domain of the function:
h(x)= 7x/[x*(x^2-9)]
----
The denominator cannot zero.
Solve: x(x^2-9) = 0
Factor:
x(x-3)(x+3) = 0
x = 0 or x = 3 or x = -3
--------------------------------
Domain is all Real Numbers except x = 0,3,or -3.
===================================
Cheers,
Stan H.
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
h(x) = 7x/[x(x^2 - 9)]
The domain will the all the numbers that we can safely replace x with that will not cause division by zero. Understand? There is no such thing as division by zero.
In the denominator of your fraction, we have two factors.
The factors are x and (x^2 - 9).
We set each factor to zero and solve for x.
The first one is easy. We get x = 0.
x^2 - 9 = 0
x^2 = 9
We take the square root of both sides.
sqrt[x^2] = sqrt[9]
Whenever we take the square root, there will be TWO answers. One answer is positive and the other is negative.
sqrt[x^2] = sqrt[9]
x = -3 and 3.
DOMAIN is: ALL REAL NUMBERS except that x CANNOT equal 0, -3 and 3.
Understand?
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
h*(x)= 7x/x*(x^2-9)
h(x)=7/(x^2-9)
=7/((x+3)(x-3))
x cannot be 3 or -3 because there would be division by zero.
The domain is all real numbers except 3 and -3.
.
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