For now we will use x 1 in both functions the first

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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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