Discover new perspectives and gain insights with IDNLearn.com. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.

Problem (1). A function [tex]\( f \)[/tex] is called even if [tex]\( f(-x)=f(x) \)[/tex] for all [tex]\( x \)[/tex]. For example, [tex]\( f(x)=x^2 \)[/tex] or [tex]\( \cos(x) \)[/tex] are even functions. A function is called odd if [tex]\( f(-x)=-f(x) \)[/tex] for all [tex]\( x \)[/tex]. For example, [tex]\( \sin(x) \)[/tex], [tex]\( \tan(x) \)[/tex], or [tex]\( f(x)=x^3 \)[/tex] are odd functions.

Suppose [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are two functions. Classify if the functions [tex]\( h(x)=f(x) g(x) \)[/tex] and [tex]\( i(x)=f(g(x)) \)[/tex] are even or odd based on whether [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are even or odd. Fill in the table below with your answer.

\begin{tabular}{c|c|c|c}
[tex]\( f \)[/tex] & [tex]\( g \)[/tex] & [tex]\( h \)[/tex] & [tex]\( i \)[/tex] \\
\hline
even & even & even & even \\
even & odd & odd & even \\
odd & even & odd & odd \\
odd & odd & even & odd \\
\end{tabular}


Sagot :

Certainly! Let's analyze the functions [tex]\( h(x) = f(x)g(x) \)[/tex] and [tex]\( i(x) = f(g(x)) \)[/tex] based on the classifications of [tex]\( f \)[/tex] and [tex]\( g \)[/tex] as either even or odd.

### Analysis of [tex]\( h(x) = f(x) g(x) \)[/tex]
#### Case 1: [tex]\( f(x) \)[/tex] is even and [tex]\( g(x) \)[/tex] is even
- For an even function [tex]\( f \)[/tex], we have [tex]\( f(-x) = f(x) \)[/tex].
- For an even function [tex]\( g \)[/tex], we have [tex]\( g(-x) = g(x) \)[/tex].
- Therefore, [tex]\( h(-x) = f(-x)g(-x) = f(x)g(x) = h(x) \)[/tex].
- Conclusion: [tex]\( h(x) \)[/tex] is even.

#### Case 2: [tex]\( f(x) \)[/tex] is even and [tex]\( g(x) \)[/tex] is odd
- For an even function [tex]\( f \)[/tex], we have [tex]\( f(-x) = f(x) \)[/tex].
- For an odd function [tex]\( g \)[/tex], we have [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( h(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -h(x) \)[/tex].
- Conclusion: [tex]\( h(x) \)[/tex] is odd.

#### Case 3: [tex]\( f(x) \)[/tex] is odd and [tex]\( g(x) \)[/tex] is even
- For an odd function [tex]\( f \)[/tex], we have [tex]\( f(-x) = -f(x) \)[/tex].
- For an even function [tex]\( g \)[/tex], we have [tex]\( g(-x) = g(x) \)[/tex].
- Therefore, [tex]\( h(-x) = f(-x)g(-x) = (-f(x))g(x) = -f(x)g(x) = -h(x) \)[/tex].
- Conclusion: [tex]\( h(x) \)[/tex] is odd.

#### Case 4: [tex]\( f(x) \)[/tex] is odd and [tex]\( g(x) \)[/tex] is odd
- For an odd function [tex]\( f \)[/tex], we have [tex]\( f(-x) = -f(x) \)[/tex].
- For an odd function [tex]\( g \)[/tex], we have [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( h(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) = h(x) \)[/tex].
- Conclusion: [tex]\( h(x) \)[/tex] is even.

### Analysis of [tex]\( i(x) = f(g(x)) \)[/tex]
#### Case 1: [tex]\( f(x) \)[/tex] is even and [tex]\( g(x) \)[/tex] is even
- For an even function [tex]\( f \)[/tex], we have [tex]\( f(-x) = f(x) \)[/tex].
- For an even function [tex]\( g \)[/tex], we have [tex]\( g(-x) = g(x) \)[/tex].
- Therefore, [tex]\( i(-x) = f(g(-x)) = f(g(x)) = i(x) \)[/tex].
- Conclusion: [tex]\( i(x) \)[/tex] is even.

#### Case 2: [tex]\( f(x) \)[/tex] is even and [tex]\( g(x) \)[/tex] is odd
- For an even function [tex]\( f \)[/tex], we have [tex]\( f(-x) = f(x) \)[/tex].
- For an odd function [tex]\( g \)[/tex], we have [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( i(-x) = f(g(-x)) = f(-g(x)) = f(g(x)) = i(x) \)[/tex].
- Conclusion: [tex]\( i(x) \)[/tex] is even.

#### Case 3: [tex]\( f(x) \)[/tex] is odd and [tex]\( g(x) \)[/tex] is even
- For an odd function [tex]\( f \)[/tex], we have [tex]\( f(-x) = -f(x) \)[/tex].
- For an even function [tex]\( g \)[/tex], we have [tex]\( g(-x) = g(x) \)[/tex].
- Therefore, [tex]\( i(-x) = f(g(-x)) = f(g(x)) = i(x) \)[/tex].
- Conclusion: [tex]\( i(x) \)[/tex] is odd.

#### Case 4: [tex]\( f(x) \)[/tex] is odd and [tex]\( g(x) \)[/tex] is odd
- For an odd function [tex]\( f \)[/tex], we have [tex]\( f(-x) = -f(x) \)[/tex].
- For an odd function [tex]\( g \)[/tex], we have [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( i(-x) = f(g(-x)) = f(-g(x)) = -f(g(x)) = -i(x) \)[/tex].
- Conclusion: [tex]\( i(x) \)[/tex] is odd.

### Summary Table:
[tex]\[ \begin{array}{c|c|c|c} f & g & h & i \\ \hline \text{even} & \text{even} & \text{even} & \text{even} \\ \text{even} & \text{odd} & \text{odd} & \text{even} \\ \text{odd} & \text{even} & \text{odd} & \text{odd} \\ \text{odd} & \text{odd} & \text{even} & \text{odd} \\ \end{array} \][/tex]