Question

Find​ (a) ​(fcircle​g)(x) and​ (b) ​(gcircle​f)(x).
​f(x)equalsStartRoot x minus 1 EndRoot​; ​g(x)equals4 x minus 5
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Part 1
​(a) ​(fcircle​g)(x)equals
  
enter your response here ​(Simplify your​ answer.)
Part 2
​(b) ​(gcircle​f)(x)equals
  
enter your response here ​(Simplify your​ answer.)

Answer 1

To find the compositions of the functions \( f \) and \( g \), we need to perform \( (f \circ g)(x) \) and \( (g \circ f)(x) \).

Given:
- \( f(x) = \sqrt{x - 1} \)
- \( g(x) = 4x - 5 \)

### Part 1: \( (f \circ g)(x) \)

To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f \):

\[
(f \circ g)(x) = f(g(x)) = f(4x - 5)
\]

Now substitute \( g(x) \) into \( f(x) \):

\[
f(4x - 5) = \sqrt{(4x - 5) - 1}
\]
\[
= \sqrt{4x - 6}
\]

Thus, we have:

\[
(f \circ g)(x) = \sqrt{4x - 6}
\]

### Part 2: \( (g \circ f)(x) \)

Next, we find \( (g \circ f)(x) \) which is \( g(f(x)) \):

\[
(g \circ f)(x) = g(f(x)) = g(\sqrt{x - 1})
\]

Now substitute \( f(x) \) into \( g(x) \):

\[
g(\sqrt{x - 1}) = 4(\sqrt{x - 1}) - 5
\]

Thus, we have:

\[
(g \circ f)(x) = 4\sqrt{x - 1} - 5
\]

### Final Answers:

- **(a)** \( (f \circ g)(x) = \sqrt{4x - 6} \)
- **(b)** \( (g \circ f)(x) = 4\sqrt{x - 1} - 5 \)