Question 85401
Given:
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{{{(x+3)/((x-2)(x+5))}}} 
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Note that this function will be undefined if a division by zero is involved. (Division by
zero is not allowed in algebra.) So if either of the factors in the denominator is equal to
zero, the entire denominator is zero and this makes the function undefined.  Now we need to
determine the values of x that will make the denominator equal to zero. Note that if 
{{{x - 2 = 0}}} then one of the factors is zero and the entire denominator becomes zero. Solve
this equation and you get {{{x = 2}}}. This tells us that x cannot equal +2 or the function
is undefined.  Similarly {{{x + 5}}} cannot equal zero. So set {{{x + 5 = 0}}} and when 
you solve that you get {{{x = -5}}}. Therefore, x cannot equal -5 or the entire denominator
will become 0 because {{{x + 5}}} will be zero. In summary, there are two rational numbers
x = +2 and x = -5 for which the function is undefined.
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Next problem:
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{{{7/(x^2 + 81)}}}
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Notice that x^2 is always a positive number or zero. Therefore, the denominator of this
function cannot be zero. (It will be at least 81 and that occurs when x = 0.) You can 
also tell that the denominator cannot be zero by setting up the equation:
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{{{x^2 + 81 = 0}}}
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Subtract 81 from both sides of this equation and you get:
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{{{x^2 = -81}}}
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But this is impossible because whenever you square a positive or a negative number the
answer is always positive. So there is no way you can multiply a number by itself and
get minus 81 as the answer. From this you can conclude that the denominator of this function
cannot equal zero. So the function is defined for all real numbers assigned to x.
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Hope this helps to clarify things for you regarding these two problems.