QuestionEvaluate $(f \circ g)(x)$ and find its domain. Given $f(x)=\frac{5}{x^{2}+36}$ and $g(x)=\sqrt{4+x}$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x)=\frac{5}{x^{}+36}$ . The function $g(x)$ is given by $g(x)=\sqrt{4+x}$
3. We need to evaluate the composite function $(f \circ g)(x)$ and find its domain in interval notation.

STEP 2

First, let's evaluate the composite function $(f \circ g)(x)$ . The composite function $(f \circ g)(x)$ is defined as $f(g(x))$ .
$(f \circ g)(x)=f(g(x))$

STEP 3

Now, plug in the given function $g(x)$ into the function $f(x)$ .
$(f \circ g)(x)=f(\sqrt{+x})$

STEP 4

Substitute $g(x)$ into $f(x)$ .
$(f \circ g)(x)=\frac{}{(\sqrt{4+x})^{2}+36}$

STEP 5

implify the expression.
$(f \circ g)(x)=\frac{5}{4+x+36}$

STEP 6

Further simplify the expression.
$(f \circ g)(x)=\frac{5}{40+x}$

STEP 7

Now, let's find the domain of the composite function $(f \circ g)(x)$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output.
The denominator of the function $(f \circ g)(x)$ cannot be zero, as division by zero is undefined. Therefore, we need to find the x-values for which the denominator is not zero.
$40+x \neq0$

STEP 8

olve the inequality to find the x-values that satisfy the condition.
$x \neq -40$

STEP 9

The domain of the function $(f \circ g)(x)$ is all real numbers except $-40$ . In interval notation, this is written as $(-\infty, -40) \cup (-40, \infty)$ .
So, the composite function $(f \circ g)(x)$ is $\frac{5}{40+x}$ and its domain in interval notation is $(-\infty, -40) \cup (-40, \infty)$ .

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