Questions on the composition of functions along with their detailed step-by-step solutions are presented. There is also a tutorial on compositions of functions that may be used. Some of the questions focus on finding the composition formula and its domain. An answer is correct when both the formula and the domain are correct. Explore applications of composition of functions for more practice.
Find the composition function \( (f \circ g)(x) \) and its domain given:
\( f(x) = x + 1 \), \( g(x) = 3x \)
The composition \( (f \circ g)(x) \) is given by:
\[ (f \circ g)(x) = f(g(x)) = g(x) + 1 = 3x + 1 \]Since the domain of both individual functions is the set of all real numbers, the composition \( (f \circ g)(x) \) also has the set of all real numbers as its domain: \( (-\infty, \infty) \).
Find the composition function \( (f \circ g)(x) \) and its domain given:
\( f(x) = x^2 + 1 \), \( g(x) = \sqrt{2x} \)
The composition \( (f \circ g)(x) \) is given by:
\[ (f \circ g)(x) = f(g(x)) = (g(x))^2 + 1 = (\sqrt{2x})^2 + 1 = 2x + 1 \]To find the domain, \( x \) must be in the domain of \( g \):
\[ 2x \geq 0 \implies x \geq 0 \]Since the domain of \( f \) is all real numbers, any output of \( g(x) \) is valid. Thus, the domain of the composition is the interval:
\[ [0, +\infty) \]Find the composition function \( (f \circ g)(x) \) and its domain given:
\( f(x) = \sqrt{-x + 1} \), \( g(x) = x^2 - 8 \)
The composition \( (f \circ g)(x) \) is given by:
\[ (f \circ g)(x) = f(g(x)) = \sqrt{-g(x) + 1} = \sqrt{-(x^2 - 8) + 1} = \sqrt{9 - x^2} \]The domain of \( g \) is all real numbers. For the composition to be defined, the expression under the square root must be non-negative:
\[ 9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3 \]Thus, the domain is given by the interval:
\[ [-3, 3] \]Find, if possible, \( (f \circ g)(-2) \) given:
\( f(x) = 2x + 1 \), \( g(x) = x^2 \)
Evaluate step-by-step:
\[ (f \circ g)(-2) = f(g(-2)) \]First find \( g(-2) \):
\[ g(-2) = (-2)^2 = 4 \]Then evaluate \( f(4) \):
\[ f(4) = 2(4) + 1 = 9 \]Find, if possible, \( (f \circ g)(0) \) given:
\( f(x) = \dfrac{1}{x + 1} \), \( g(x) = \dfrac{1}{x - 1} \)
Evaluate step-by-step:
\[ (f \circ g)(0) = f(g(0)) \]First find \( g(0) \):
\[ g(0) = \dfrac{1}{0 - 1} = -1 \]Then evaluate \( f(-1) \):
\[ f(-1) = \dfrac{1}{-1 + 1} = \dfrac{1}{0} \]Since division by zero is undefined, \( (f \circ g)(0) \) is undefined (because \( -1 \) is not in the domain of \( f \)).