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Suppose that the functions \( h \) and \( g \) are defined as follows. \[ \begin{array}{l}h(x)=(x-1)(x-4) \\ g(x)=x+7 \\ \text { (a) Find }\left(\frac{h}{g}\right)(-6) \text {. } \\ \text { (b) Find all values that are NOT in the domain of } \frac{h}{g} \text {. } \\ \text { If there is more than one value, separate them with commas. } \\ \text { (a) }\left(\frac{h}{g}\right)(-6)=\square \\ \text { (b) Value(s) that are NOT in the domain of } \frac{h}{g}: \square\end{array} \] (b)

Ask by Martinez Weaver. in the United States
Mar 20,2025

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Answer

The value not in the domain of \(\frac{h}{g}\) is \(-7\).

Solution

1. The function is given by \[ g(x) = x + 7. \] 2. When finding the domain of \(\frac{h}{g}\), we need to exclude values of \(x\) for which the denominator is zero. 3. Set the denominator equal to zero: \[ x + 7 = 0. \] 4. Solve for \(x\): \[ x = -7. \] 5. Therefore, the value that is not in the domain of \(\frac{h}{g}\) is: \[ -7. \]

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To find \(\left(\frac{h}{g}\right)(-6)\), we first need to calculate \(h(-6)\) and \(g(-6)\). 1. Calculate \(h(-6)\): \[ h(-6) = (-6 - 1)(-6 - 4) = (-7)(-10) = 70 \] 2. Calculate \(g(-6)\): \[ g(-6) = -6 + 7 = 1 \] Now we can find \(\left(\frac{h}{g}\right)(-6)\): \[ \left(\frac{h}{g}\right)(-6) = \frac{h(-6)}{g(-6)} = \frac{70}{1} = 70 \] Next, to find the values that are NOT in the domain of \(\frac{h}{g}\), we must identify when \(g(x) = 0\) because division by zero is undefined. Setting \(g(x) = 0\): \[ x + 7 = 0 \Rightarrow x = -7 \] So the only value that is NOT in the domain of \(\frac{h}{g}\) is \(-7\). Thus, the final answers are: (a) \(\left(\frac{h}{g}\right)(-6) = 70\) (b) Value(s) that are NOT in the domain of \(\frac{h}{g}: -7\)

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