Prove that f(x)=|x| is strictly increasing function in the interval ]0,∞[ and strictly decreasing in the ]-∞,0[

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!

Prove that f(x) = |x| is strictly increasing in the interval ]0,∞[ and strictly decreasing in ]-∞,0[

Explore how to prove that the absolute value function f(x) = |x| is strictly increasing on the interval ]0, ∞[ and strictly decreasing on ]-∞, 0[ using derivatives. This problem covers key concepts in calculus, including monotonicity and differentiability. Suitable for Grades 10-12.

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