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Suppose that the functions [tex]\( q \)[/tex] and [tex]\( r \)[/tex] are defined as follows:

[tex]\[
\begin{array}{l}
q(x) = x^2 + 7 \\
r(x) = \sqrt{x + 4}
\end{array}
\][/tex]

Find the following:

[tex]\[
\begin{array}{l}
(q \circ r)(5) = \\
(r \circ q)(5) =
\end{array}
\][/tex]


Sagot :

Alright, we are given the functions [tex]\( q(x) = x^2 + 7 \)[/tex] and [tex]\( r(x) = \sqrt{x+4} \)[/tex]. We need to find [tex]\((q \circ r)(5)\)[/tex] and [tex]\((r \circ q)(5)\)[/tex].

Let's start with [tex]\((q \circ r)(5)\)[/tex]:

1. First, we evaluate [tex]\( r(5) \)[/tex]:
[tex]\[ r(5) = \sqrt{5 + 4} = \sqrt{9} = 3 \][/tex]

2. Then, we use this result as the input for [tex]\( q \)[/tex]:
[tex]\[ q(r(5)) = q(3) \][/tex]
Now, we compute [tex]\( q(3) \)[/tex]:
[tex]\[ q(3) = 3^2 + 7 = 9 + 7 = 16 \][/tex]

Thus, [tex]\((q \circ r)(5) = 16\)[/tex].

Next, we find [tex]\((r \circ q)(5)\)[/tex]:

1. First, we evaluate [tex]\( q(5) \)[/tex]:
[tex]\[ q(5) = 5^2 + 7 = 25 + 7 = 32 \][/tex]

2. Then, we use this result as the input for [tex]\( r \)[/tex]:
[tex]\[ r(q(5)) = r(32) \][/tex]
Now, we compute [tex]\( r(32) \)[/tex]:
[tex]\[ r(32) = \sqrt{32 + 4} = \sqrt{36} = 6 \][/tex]

Thus, [tex]\((r \circ q)(5) = 6\)[/tex].

Hence, the solutions are:
[tex]\[ (q \circ r)(5) = 16 \][/tex]
[tex]\[ (r \circ q)(5) = 6 \][/tex]
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