SOLUTION: Find the value of a which will ensure that the domain of the function f(x) =3/sqrt(ax+1) ----------- ----------- ----------- ----------- ----------- ----------- --------

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Question 73849: Find the value of a which will ensure that the domain of the function
f(x) =3/sqrt(ax+1)
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is the interval (-¥, 10), i.e. x < 10.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The domain is the set of numbers that are allowed in the function. For instance, for the function 1%2Fx, if we plug in x=0 we would get an error since we cannot divide by zero. So x=0 is not in our domain. So we want a value of a that will give us a domain of (-∞,10). To find this value, we need to specify that we cannot get 0 in our denominator or get a negative value in our square root. If we let x=10 we get
3%2Fsqrt%2810a%2B1%29
So we can say
10a%2B1%3E0So we can avoid dividing by 0
10a%3E-1
a%3E-1%2F10
So a must equal -1/10 to have a domain of (-∞10) since if we let x=10
3%2Fsqrt%2810%28-1%2F10%29%2B1%29
3%2Fsqrt%28-1%2B1%29
3%2Fsqrt%280%29
3%2F0Which is not possible, so it shows that everything less than 10 will work. For instance
3%2Fsqrt%288%28-1%2F10%29%2B1%29
3%2Fsqrt%28-8%2F10%2B1%29
3%2Fsqrt%280.2%29Which can be done since the denominator is greater than zero.