Math

QuestionFind (qr)(x)(q \circ r)(x) for q(x)=1x10q(x)=\frac{1}{x-10} and r(x)=4x5r(x)=|4x-5|. What is the domain in interval notation?

Studdy Solution

STEP 1

Assumptions1. We have two functions, q(x)q(x) and r(x)r(x). . q(x)=1x10q(x) = \frac{1}{x-10}
3. r(x)=4x5r(x) = |4x-5|
4. We are asked to evaluate the composite function (qr)(x)(q \circ r)(x) and find its domain.

STEP 2

First, let's understand what a composite function is. In mathematics, the composition of two functions, qq and rr, denoted as (qr)(x)(q \circ r)(x), is a function that applies rr to its input, and then applies qq to the result.

STEP 3

Now, let's substitute r(x)r(x) into q(x)q(x) to find the composite function (qr)(x)(q \circ r)(x).
(qr)(x)=q(r(x))(q \circ r)(x) = q(r(x))

STEP 4

Substitute r(x)=4xr(x) = |4x-| into q(x)=1x10q(x) = \frac{1}{x-10}.
(qr)(x)=q(4x)=14x10(q \circ r)(x) = q(|4x-|) = \frac{1}{|4x-|-10}

STEP 5

Now, let's find the domain of (qr)(x)(q \circ r)(x). The domain of a function is the set of all possible inputs, or in other words, the set of all real numbers xx for which the function is defined.

STEP 6

The function (qr)(x)=14x510(q \circ r)(x) = \frac{1}{|4x-5|-10} is undefined when the denominator equals zero. So, we need to find the values of xx for which 4x510=0|4x-5|-10 =0.

STEP 7

olve the equation 4x510=0|4x-5|-10 =0 for xx.
4x5=10|4x-5| =10

STEP 8

The absolute value equation 4x5=10|4x-5| =10 splits into two equations 4x5=104x-5 =10 and 4x5=104x-5 = -10.

STEP 9

olve the equation 4x5=4x-5 = for xx.
4x=154x =15x=154x = \frac{15}{4}

STEP 10

olve the equation 4x5=104x-5 = -10 for xx.
4x=54x = -5x=54x = -\frac{5}{4}

STEP 11

The function (qr)(x)(q \circ r)(x) is undefined at x=154x = \frac{15}{4} and x=54x = -\frac{5}{4}. Therefore, the domain of (qr)(x)(q \circ r)(x) in interval notation is (,54)(54,154)(154,)(-\infty, -\frac{5}{4}) \cup (-\frac{5}{4}, \frac{15}{4}) \cup (\frac{15}{4}, \infty).
So, the composite function (qr)(x)=4x510(q \circ r)(x) = \frac{}{|4x-5|-10} and its domain is (,54)(54,154)(154,)(-\infty, -\frac{5}{4}) \cup (-\frac{5}{4}, \frac{15}{4}) \cup (\frac{15}{4}, \infty).

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