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# Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for

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Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

Guest Sep 15, 2014

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Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

$$\\Let \displaystyle f(x) = \frac{1}{x-3} \quad and \quad \displaystyle g(x) = \frac{1}{x-7}. \\\\ \mbox{Then the domain of f\circ g is equal to all reals except for two values, a and b with a }$$

$$\\f \circ g=\dfrac{1}{\frac{1}{x-7}-3}\\\\ =\dfrac{1}{\frac{1-3(x-7)}{x-7}}\\\\\\ =\dfrac{1}{\frac{-3x+22}{x-7}}\\\\\\ =1 \div \frac{-3x+22}{x-7}\\\\ =1 \times \frac{x-7}{-3x+22}\\\\ =\frac{x-7}{22-3x}\\\\$$

NOW     x cannot equal to 7 or 22/3  because you cannot divide by zero!

$$x \in R \quad where \;x\ne7 \;and\;x\ne \frac{22}{3}$$

Melody  Sep 16, 2014
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#1
+91436
+13

Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

$$\\Let \displaystyle f(x) = \frac{1}{x-3} \quad and \quad \displaystyle g(x) = \frac{1}{x-7}. \\\\ \mbox{Then the domain of f\circ g is equal to all reals except for two values, a and b with a }$$

$$\\f \circ g=\dfrac{1}{\frac{1}{x-7}-3}\\\\ =\dfrac{1}{\frac{1-3(x-7)}{x-7}}\\\\\\ =\dfrac{1}{\frac{-3x+22}{x-7}}\\\\\\ =1 \div \frac{-3x+22}{x-7}\\\\ =1 \times \frac{x-7}{-3x+22}\\\\ =\frac{x-7}{22-3x}\\\\$$

NOW     x cannot equal to 7 or 22/3  because you cannot divide by zero!

$$x \in R \quad where \;x\ne7 \;and\;x\ne \frac{22}{3}$$

Melody  Sep 16, 2014

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