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

#1**+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

#1**+13 **

Best Answer

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