Let's go through the exercises in the image you provided one by one.

Exercise 3.2.35:

Problem: Show that $f(x) = x^2$ is decreasing in $(-\infty, 0]$ and increasing in $[0, \infty)$.

Solution Strategy:

• Differentiate $f(x) = x^2$ to find the derivative: $f'(x) = 2x$
• For $x < 0$, $f'(x) = 2x$ is negative, so $f(x)$ is decreasing.
• For $x > 0$, $f'(x) = 2x$ is positive, so $f(x)$ is increasing.
• At $x = 0$, the derivative $f'(0) = 0$, but it marks a turning point (minimum), so the function transitions from decreasing to increasing.

Exercise 3.2.36:

Problem: Let $f: I \to \mathbb{R}$ be strictly increasing. Then show that $f$ is injective. What about strictly decreasing functions?

Solution Strategy:

• Strictly Increasing Case:
  A function is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Since $f$ is strictly increasing, if $x_1 < x_2$, then $f(x_1) < f(x_2)$. Therefore, it cannot happen that $f(x_1) = f(x_2)$ for distinct values of $x_1$ and $x_2$, which proves injectivity.

• Strictly Decreasing Case:
  Similarly, if $f$ is strictly decreasing, for $x_1 < x_2$, we have $f(x_1) > f(x_2)$. Thus, it is impossible for $f(x_1) = f(x_2)$ if $x_1 \neq x_2$, so strictly decreasing functions are also injective.

Exercise 3.2.37:

Problem: Let $a, b \in \mathbb{R}$ and $f(x) = ax + b$ for $x \in \mathbb{R}$. Show that $f$ is strictly increasing if $a > 0$ and strictly decreasing if $a < 0$. When is $f$ injective?

Solution Strategy:

• Differentiate $f(x) = ax + b$: $f'(x) = a$
• If $a > 0$, the slope $f'(x) = a$ is positive, so $f$ is strictly increasing.
• If $a < 0$, the slope $f'(x) = a$ is negative, so $f$ is strictly decreasing.
• Since both strictly increasing and strictly decreasing functions are injective (as shown in Exercise 3.2.36), $f$ is injective for all $a \neq 0$.

Exercise 3.2.39:

Problem: Let $f: I \to \mathbb{R}$ be both decreasing and increasing. Prove that $f$ is a constant function.

Solution Strategy:

• If $f$ is both decreasing and increasing, it means that for any two points $x_1, x_2 \in I$, the values $f(x_1)$ and $f(x_2)$ must be equal.
• A function that neither strictly increases nor decreases over any interval is constant. Therefore, $f(x) = c$ for some constant $c \in \mathbb{R}$ over the entire interval $I$.

Would you like a more detailed explanation of any of these solutions? Here are 5 follow-up questions to extend your understanding:

1. Can you explain why a function that is both increasing and decreasing must be constant?
2. How would the conclusions change if the function $f(x) = x^2$ were defined only for $x \in [0, \infty)$?
3. Can you give an example of a function that is neither injective nor strictly monotonic?
4. What additional conditions could make a function that is neither strictly increasing nor strictly decreasing still injective?
5. How does the behavior of linear functions change if the slope is zero (i.e., $a = 0$)?

Tip: When working with injective functions, remember that they must map distinct inputs to distinct outputs, which is often checked using derivatives or monotonicity properties.