Tutorial on Composition of Functions

By definition of composite functions:

\[ (f \circ g)(-2) = f(g(-2)) \]

First, evaluate \( g(-2) \):

\[ g(-2) = |-2 - 4| = |-6| = 6 \]

Next, evaluate \( f(g(-2)) = f(6) \):

\[ f(6) = -3(6) + 2 = -18 + 2 = -16 \]

Conclusion: \( (f \circ g)(-2) = -16 \)

First, find \( (f \circ g)(x) \):

\[ (f \circ g)(x) = f(g(x)) = \dfrac{g(x) - 1}{g(x) + 2} \] \[ = \dfrac{\dfrac{x + 1}{x - 2} - 1}{\dfrac{x + 1}{x - 2} + 2} = \dfrac{\dfrac{(x + 1) - (x - 2)}{x - 2}}{\dfrac{(x + 1) + 2(x - 2)}{x - 2}} = \dfrac{3}{3x - 3} = \dfrac{1}{x - 1} \]

Next, determine the domain:

• Domain of \( f \): \( x \neq -2 \)
• Domain of \( g \): \( x \neq 2 \)
• Condition for composition: \( g(x) \) must be in the domain of \( f \), meaning \( g(x) \neq -2 \).

Solve \( g(x) = -2 \):

\[ \dfrac{x + 1}{x - 2} = -2 \implies x + 1 = -2(x - 2) \implies x + 1 = -2x + 4 \implies 3x = 3 \implies x = 1 \]

Thus, \( x \) must also be different from \( 1 \).

Conclusion: The domain of \( f \circ g \) is:

\[ (-\infty, 1) \cup (1, 2) \cup (2, \infty) \]

First, find \( (f \circ g)(x) \):

\[ (f \circ g)(x) = f(g(x)) = (g(x))^2 + 2 \] \[ = (\sqrt{x - 2})^2 + 2 = x - 2 + 2 = x \]

Next, determine the domain:

• Domain of \( f \): All real numbers \( (-\infty, \infty) \)
• Domain of \( g \): \( x - 2 \geq 0 \implies x \geq 2 \)

Since the domain of \( f \) is all real numbers, we must ensure \( x \) is in the domain of \( g \) so that \( g(x) \) yields a real value.

Conclusion: The domain of \( f \circ g \) is:

\[ [2, \infty) \]

1. \( (f \circ g)(x) = 2(x^2)^3 + x^2 - 1 = 2x^6 + x^2 - 1 \),   Domain: \( (-\infty, +\infty) \)
2. \( (f \circ g)(x) = |(x-1)^2 - 4| = |x^2 - 2x + 1 - 4| = |x^2 - 2x - 3| \),   Domain: \( (-\infty, +\infty) \)
3. \( (f \circ g)(x) = (\sqrt{x+5})^2 - 5 = x + 5 - 5 = x \),   Domain: \( [-5, +\infty) \)
4. \( (f \circ g)(x) = \ln(\sqrt{x+5}) = \dfrac{1}{2}\ln(x+5) \),   Domain: \( [-5, +\infty) \)
5. \( (f \circ g)(x) = \sin(x - 2) \),   Domain: \( (-\infty, +\infty) \)