Question

try
identify the rules for ( f circ g ) and ( g circ f )
a. ( f(x) = x^3 ), ( g(x) = x + 1 )
b. ( f(x) = x + sqrt{7} ), ( g(x) = 2x - 5 )

Explanation:

Part a:

Step1: Find \( f \circ g \)

To find \( f \circ g \), we substitute \( g(x) \) into \( f(x) \). Given \( f(x) = x^3 \) and \( g(x) = x + 1 \), we replace \( x \) in \( f(x) \) with \( g(x) \).
\( f(g(x)) = (g(x))^3 = (x + 1)^3 \)
We can expand \( (x + 1)^3 \) using the formula \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \), where \( a = x \) and \( b = 1 \).
\( (x + 1)^3 = x^3 + 3x^2(1) + 3x(1)^2 + 1^3 = x^3 + 3x^2 + 3x + 1 \)

Step2: Find \( g \circ f \)

To find \( g \circ f \), we substitute \( f(x) \) into \( g(x) \). Given \( g(x) = x + 1 \) and \( f(x) = x^3 \), we replace \( x \) in \( g(x) \) with \( f(x) \).
\( g(f(x)) = f(x) + 1 = x^3 + 1 \)

Step1: Find \( f \circ g \)

To find \( f \circ g \), we substitute \( g(x) \) into \( f(x) \). Given \( f(x) = x + \sqrt{7} \) and \( g(x) = 2x - 5 \), we replace \( x \) in \( f(x) \) with \( g(x) \).
\( f(g(x)) = g(x) + \sqrt{7} = (2x - 5) + \sqrt{7} = 2x - 5 + \sqrt{7} \)

Step2: Find \( g \circ f \)

To find \( g \circ f \), we substitute \( f(x) \) into \( g(x) \). Given \( g(x) = 2x - 5 \) and \( f(x) = x + \sqrt{7} \), we replace \( x \) in \( g(x) \) with \( f(x) \).
\( g(f(x)) = 2(f(x)) - 5 = 2(x + \sqrt{7}) - 5 = 2x + 2\sqrt{7} - 5 \)

Answer:

(Part a):
\( f \circ g(x) = (x + 1)^3 = x^3 + 3x^2 + 3x + 1 \)
\( g \circ f(x) = x^3 + 1 \)

Part b: