Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
0. Functions
Properties of Functions
Problem 1.17
Textbook Question
Suppose that ƒ and g are both odd functions defined on the entire real line. Which of the following (where defined) are even? odd?
a. ƒg
b. ƒ³
c. ƒ(sin x)
d. g(sec x)
e. |g|

1
Recall that a function \( h(x) \) is odd if \( h(-x) = -h(x) \) for all \( x \), and it is even if \( h(-x) = h(x) \) for all \( x \).
For part (a), consider the product \( f(x)g(x) \). Since both \( f \) and \( g \) are odd, \( f(-x) = -f(x) \) and \( g(-x) = -g(x) \). Therefore, \( f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) \), making \( f(x)g(x) \) an even function.
For part (b), consider \( f^3(x) = (f(x))^3 \). Since \( f \) is odd, \( f(-x) = -f(x) \), so \( f^3(-x) = (-f(x))^3 = -f^3(x) \). Thus, \( f^3(x) \) is odd.
For part (c), consider \( f(\sin x) \). Since \( \sin(-x) = -\sin(x) \), and \( f \) is odd, \( f(\sin(-x)) = f(-\sin(x)) = -f(\sin(x)) \). Therefore, \( f(\sin x) \) is odd.
For part (d), consider \( g(\sec x) \). Since \( \sec(-x) = \sec(x) \), and \( g \) is odd, \( g(\sec(-x)) = g(\sec(x)) = -g(\sec(x)) \). This is a contradiction unless \( g(\sec x) = 0 \), so \( g(\sec x) \) is not defined as an odd or even function. For part (e), consider \( |g(x)| \). The absolute value function makes \( |g(x)| \) even because \( |g(-x)| = |g(x)| \).
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