QuestionFind $(f \circ g)(x)$ and its domain. Given $f(x)=\frac{x}{x-1}$ and $g(x)=\frac{13}{x^{2}-36}$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x)=\frac{x}{x-1}$ . The function $g(x)$ is given by $g(x)=\frac{13}{x^{}-36}$
3. We are asked to evaluate the composition of functions $(f \circ g)(x)$ and find its domain in interval notation.

STEP 2

The composition of functions $(f \circ g)(x)$ is defined as $f(g(x))$ .

STEP 3

Substitute $g(x)$ into $f(x)$ to find $(f \circ g)(x)$ .
$(f \circ g)(x) = f(g(x)) = f\left(\frac{13}{x^{2}-36}\right)$

STEP 4

Now, replace $x$ in $f(x)$ with $g(x)$ to get the expression for $(f \circ g)(x)$ .
$(f \circ g)(x) = \frac{\frac{13}{x^{2}-36}}{\frac{13}{x^{2}-36}-1}$

STEP 5

implify the expression to get $(f \circ g)(x)$ .
$(f \circ g)(x) = \frac{13}{49-x^{2}}$

STEP 6

Now, let's find the domain of $(f \circ g)(x)$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output.

STEP 7

The denominator of the function $(f \circ g)(x)$ cannot be zero, as division by zero is undefined. So, we set the denominator equal to zero and solve for $x$ .
$49-x^{2} =0$

STEP 8

olve the equation to find the values of $x$ that are not in the domain.
$x = \pm \sqrt{49} = \pm7$

STEP 9

So, the domain of $(f \circ g)(x)$ is all real numbers except $-7$ and $7$ .

STEP 10

In interval notation, this is written as $(-\infty, -7) \cup (-7,7) \cup (7, \infty)$ .
So, the function $(f \circ g)(x) = \frac{13}{49-x^{2}}$ and its domain in interval notation is $(-\infty, -7) \cup (-7,7) \cup (7, \infty)$ .

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