George Morgan
06/22/2023 · Middle School
For the functions \( f(x)=-10 x+15 \) and \( g(x)=5 x+7 \) find the following. (a) \( (f \circ g)(x) \) and its domain (b) \( (g \circ f)(x) \) and its domain (a) \( (f \circ g)(x)=-50 x-55 \) (Simplify your answer.) The domain is (Type your answer in interval notation.)
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Step-by-step Solution
To find the compositions of the functions \( f(x) = -10x + 15 \) and \( g(x) = 5x + 7 \), we will calculate \( (f \circ g)(x) \) and \( (g \circ f)(x) \).
### (a) Finding \( (f \circ g)(x) \)
The composition \( (f \circ g)(x) \) means we will substitute \( g(x) \) into \( f(x) \):
\[
g(x) = 5x + 7
\]
Now substitute \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f(5x + 7) = -10(5x + 7) + 15
\]
Now simplify:
\[
= -50x - 70 + 15
\]
\[
= -50x - 55
\]
Thus, we have:
\[
(f \circ g)(x) = -50x - 55
\]
**Domain of \( (f \circ g)(x) \)**: Since both \( f(x) \) and \( g(x) \) are linear functions, their domains are all real numbers. Therefore, the domain of \( (f \circ g)(x) \) is:
\[
\text{Domain: } (-\infty, \infty)
\]
### (b) Finding \( (g \circ f)(x) \)
Now we will find \( (g \circ f)(x) \), which means we will substitute \( f(x) \) into \( g(x) \):
\[
f(x) = -10x + 15
\]
Now substitute \( f(x) \) into \( g(x) \):
\[
g(f(x)) = g(-10x + 15) = 5(-10x + 15) + 7
\]
Now simplify:
\[
= -50x + 75 + 7
\]
\[
= -50x + 82
\]
Thus, we have:
\[
(g \circ f)(x) = -50x + 82
\]
**Domain of \( (g \circ f)(x) \)**: Similar to the previous case, since both \( g(x) \) and \( f(x) \) are linear functions, their domains are all real numbers. Therefore, the domain of \( (g \circ f)(x) \) is:
\[
\text{Domain: } (-\infty, \infty)
\]
### Summary of Results
- \( (f \circ g)(x) = -50x - 55 \) with domain \( (-\infty, \infty) \)
- \( (g \circ f)(x) = -50x + 82 \) with domain \( (-\infty, \infty) \)
Quick Answer
(a) \( (f \circ g)(x) = -50x - 55 \) with domain \( (-\infty, \infty) \)
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