Question 1040542: I need help solving this problem please!
f(x) = sqrt(x+5) and g(x)=x-3
I need to find (f/g)(-1) and I'm coming up with:
sqrt((x+5)(x-3))/ x-3
I'm also asked to find the domain and come up with:
(−∞,−5]∪(3,∞)
am I on the right track?
Thank you for any guidance!
Found 2 solutions by stanbon, Aldorozos:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! f(x) = sqrt(x+5) and g(x)=x-3
I need to find (f/g)(-1) and I'm coming up with:
f(-1) = sqrt(-1+5) = sqrt(4) = 2
g(-1) = -1-3 = -4
(f/g)(-1) = 2/-4 = -1/2
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(f/g)(x) = sqrt(x+5)/(x-3)
I'm also asked to find the domain and come up with:
Since x cannot be "3",
Domain:: (-oo,3)U(3,+oo)
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Cheers,
Stan H.
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Answer by Aldorozos(172) (Show Source):
You can put this solution on YOUR website! First you have to divide both functions
Sqrt(x+5)/x-3 now you have to replace x with -1
Sqrt(-1+5)/(-1-3)= sqrt4/-4 2/-4= -1/2
We know that sqrt can not be less than zero. In other words sqrt is always zero or greater than zero
We also know that the denominator can't be zero
Now we look at
Sqrt(x+5) this has to be greater or equal to zero. This means that x has to be -5 or greater than -5
We said that the denominator can't be zero. This means that x-3 can't be zero the only time x-3 is zero is when x is 3. Therefore x can't be 3
We said x has to be -5 or greater and it can't be 3
The way we can rewrite this statement is
[-5,3) and (3,infinitie)
I think the statement you have written is not accurate when you said
(Infiniti,-5]. By saying that you are saying the numbers less than -5 are acceptable. You can try -6 for example and put it under the square root. You will get sqrt(-6+5)=sqrt of -1 we know that we don't have any real number that if multipiled by itself could give us -1. In other words we can't have a nevative number under sqrt. This means the number has to be -5 or greater than -5. Also that nber can't be 3.
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