To prove that $f(x) = |x|$ is strictly increasing in the interval $]0, \infty[$ and strictly decreasing in the interval $]-\infty, 0[$, we will approach this by analyzing the behavior of the function on each interval separately. The key to this is understanding the derivative of $f(x)$ and how it behaves across the given intervals.

Definition of $f(x) = |x|$:

• When $x \geq 0$, $f(x) = |x| = x$.
• When $x < 0$, $f(x) = |x| = -x$.

Case 1: $f(x) = |x|$ on $]0, \infty[$

In this interval, $f(x) = x$. To prove that it is strictly increasing, we compute the derivative of $f(x)$.

$f'(x) = \frac{d}{dx}(x) = 1$

Since $f'(x) = 1$ for all $x > 0$, the function is strictly increasing in $]0, \infty[$, because the derivative is positive throughout this interval.

Case 2: $f(x) = |x|$ on $]-\infty, 0[$

In this interval, $f(x) = -x$. To prove that it is strictly decreasing, we compute the derivative of $f(x)$.

$f'(x) = \frac{d}{dx}(-x) = -1$

Since $f'(x) = -1$ for all $x < 0$, the function is strictly decreasing in $]-\infty, 0[$, because the derivative is negative throughout this interval.

Conclusion:

• $f(x) = |x|$ is strictly increasing on $]0, \infty[$ because $f'(x) = 1 > 0$.
• $f(x) = |x|$ is strictly decreasing on $]-\infty, 0[$ because $f'(x) = -1 < 0$.

This confirms that $f(x) = |x|$ behaves as described in the two intervals.

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Would you like more details on the behavior of absolute value functions, or do you have any other questions?

Here are five related questions to explore further:

1. How does the derivative test work for determining increasing and decreasing functions?
2. Can a function be non-differentiable at a point but still be continuous?
3. How does the absolute value function behave at $x = 0$?
4. What are examples of other piecewise-defined functions similar to $f(x) = |x|$?
5. How do we prove the concavity of a function on given intervals?

Tip: The absolute value function has a cusp at $x = 0$, making it non-differentiable at that point!