Question

for the real - valued functions ( g(x)=\frac{x + 6}{x+3} ) and ( h(x)=4x + 1 ), find the composition ( gcirc h ) and specify its domain using interval notation.
( (gcirc h)(x)=quad )
domain of ( gcirc h ): (quad)

Explanation:

Step1: Find \( (g \circ h)(x) \)

To find the composition \( g \circ h \), we substitute \( h(x) \) into \( g(x) \). Given \( g(x) = \frac{x + 6}{x + 3} \) and \( h(x) = 4x + 1 \), we replace \( x \) in \( g(x) \) with \( 4x + 1 \):

$$ (g \circ h)(x) = g(h(x)) = \frac{(4x + 1) + 6}{(4x + 1) + 3} $$

Simplify the numerator and the denominator:

$$ (g \circ h)(x) = \frac{4x + 7}{4x + 4} $$

Step2: Find the domain of \( g \circ h \)

The domain of a composition \( g \circ h \) is the set of all \( x \) such that \( x \) is in the domain of \( h \) and \( h(x) \) is in the domain of \( g \).

First, the domain of \( h(x) = 4x + 1 \) (a linear function) is all real numbers, \( (-\infty, \infty) \).

Next, the domain of \( g(x) = \frac{x + 6}{x + 3} \) excludes values where the denominator is zero. So, we set the denominator of \( g \) equal to zero and solve for \( x \):

$$ x + 3 = 0 \implies x = -3 $$

Thus, \( h(x) \) must not equal \( -3 \). We set \( h(x) = -3 \) and solve for \( x \):

$$ 4x + 1 = -3 $$

Subtract 1 from both sides:

$$ 4x = -4 $$

Divide by 4:

$$ x = -1 $$

So, \( x = -1 \) must be excluded from the domain of \( g \circ h \). Since the domain of \( h \) is all real numbers, the domain of \( g \circ h \) is all real numbers except \( x = -1 \).

In interval notation, this is \( (-\infty, -1) \cup (-1, \infty) \).

Answer:

\( (g \circ h)(x) = \frac{4x + 7}{4x + 4} \)

Domain of \( g \circ h \): \( (-\infty, -1) \cup (-1, \infty) \)