Let's address both of the questions in the image.

Problem 3:

You are asked whether the statement is true or false:

Given:

• $f(x)$ is an even function, meaning $f(-x) = f(x)$ for all real numbers $x$.
• The function $g(x)$ is defined piecewise: \begin{cases} f(x) & \text{if } x \geq 0 \\ -f(x) & \text{if } x < 0 \end{cases}$$
• You need to determine whether $g(x)$ is an odd function.

Explanation:

To check if $g(x)$ is an odd function, we need to verify whether $g(-x) = -g(x)$ for all $x$.

Let's consider two cases:

1. When $x \geq 0$:

   • In this case, $g(x) = f(x)$.
   • We need to compute $g(-x)$. Since $-x < 0$ when $x \geq 0$, we use the second part of the piecewise definition: $g(-x) = -f(-x)$
   • Because $f(x)$ is even, we know $f(-x) = f(x)$. Hence, $g(-x) = -f(x)$
   • Therefore, $g(-x) = -g(x)$, which satisfies the condition for $g(x)$ to be odd.

2. When $x < 0$:

   • In this case, $g(x) = -f(x)$.
   • We compute $g(-x)$. Since $-x \geq 0$, we use the first part of the piecewise definition: $g(-x) = f(-x)$
   • Since $f(x)$ is even, $f(-x) = f(x)$. Therefore, $g(-x) = f(x)$
   • We now check $-g(x)$ in this case. Since $g(x) = -f(x)$, $-g(x) = -(-f(x)) = f(x)$
   • Thus, $g(-x) = -g(x)$, which again satisfies the condition for $g(x)$ to be odd.

Conclusion: Since $g(-x) = -g(x)$ holds in both cases, the statement is true: $g(x)$ is an odd function.

Problem 4:

You are asked to show that $f \cdot g = g \cdot f$ if and only if $d(a - 1) = b(c - 1)$, where:

• $f(x) = ax + b$
• $g(x) = cx + d$

Solution:

To show this, we need to compute $f(g(x))$ and $g(f(x))$, then compare the two expressions and solve for the given condition.

1. Compute $f(g(x))$: $f(g(x)) = f(cx + d) = a(cx + d) + b = acx + ad + b$

2. Compute $g(f(x))$: $g(f(x)) = g(ax + b) = c(ax + b) + d = acx + bc + d$

3. Set the two equal: $acx + ad + b = acx + bc + d$

   • Cancel out $acx$ from both sides: $ad + b = bc + d$
   • Rearranging this equation: $ad - d = bc - b$
   • Factor both sides: $d(a - 1) = b(c - 1)$

Thus, the condition $d(a - 1) = b(c - 1)$ must hold for $f \cdot g = g \cdot f$.

Do you need more details or have any questions?

Here are 5 related questions for practice:

1. How would the result of Problem 3 change if $f(x)$ were odd instead of even?
2. Can you find another example where a piecewise-defined function turns an even function into an odd function?
3. What happens if you change the conditions of Problem 4 by making both functions quadratic instead of linear?
4. How do you prove that two functions commute under composition in general?
5. Can you derive another condition involving $a$, $b$, $c$, and $d$ where $f(g(x)) \neq g(f(x))$?

Tip: When working with piecewise functions, always check conditions for different intervals separately to ensure all cases are covered properly.