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Unformatted text preview: oblems] (c) h ( x ) = 7x + 8
x2 + 4 In this case we’ve got a fraction, but notice that the denominator will never be zero for any real
number since x2 is guaranteed to be positive or zero and adding 4 onto this will mean that the
denominator is always at least 4. In other words, the denominator won’t ever be zero. So, all we
need to do then is worry about the square root in the numerator.
To do this we’ll require, 7x + 8 ³ 0
7 x ³ 8
8
x³7 Now, we can actually plug in any value of x into the denominator, however, since we’ve got the
square root in the numerator we’ll have to make sure that all x’s satisfy the inequality above to
avoid problems. Therefore, the domain of this function is Domain : x ³  8
7
[Return to Problems] (d) R ( x ) = 10 x  5
x 2  16 In this final part we’ve got both a square root and division by zero to worry about. Let’s take care
of the square root first since this will probably put the largest restriction on the values of x. So, to
keep the square roo...
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 Spring '12
 MrVinh

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