Custom Admonitions

Custom Admonitions#

The custom admonition directives provide styled content blocks for educational materials.

Theory#

For presenting theoretical content:

Pythagoras’ Theorem

For a right triangle with sides \(a\) and \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

This fundamental relation has been known since ancient times.

Example#

For worked examples:

Solving a Quadratic Equation

Solve: \(x^2 - 5x + 6 = 0\)

Using the quadratic formula: $\(x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}\)$

So \(x = 3\) or \(x = 2\)

Exercise#

For student exercises:

Practice Problem

Find the derivative of \(f(x) = x^3 + 2x^2 - 5x + 1\).

Explore#

For exploratory activities:

Investigation

Try different values and observe the pattern:

  • What happens when \(x = 0\)?

  • What happens when \(x\) is very large?

  • Can you find a general rule?

Goals#

For learning objectives and goals:

Learning Objectives

After this section, you should be able to:

  • Understand the concept of derivatives

  • Apply the chain rule

  • Solve optimization problems

Summary#

For section summaries:

Key Points

  • Derivatives measure rate of change

  • Integration finds area under curves

  • The fundamental theorem connects them

Hints#

For providing hints (default collapsed):

Hint

Start by factoring out the common term.

Custom Hint Title

This hint is always visible.

Answer#

For short answers (default collapsed):

Fasit

\(x = 42\)

Answer to Question 2

The solution is \(y = 2x + 3\).

Solution#

For full solutions (default collapsed):

Løsning

Step 1: Write down what we know

  • Initial velocity: \(v_0 = 0\)

  • Acceleration: \(a = 9.8\) m/s²

  • Time: \(t = 3\) s

Step 2: Apply the formula $\(v = v_0 + at = 0 + 9.8 \times 3 = 29.4 \text{ m/s}\)$

Answer: The final velocity is 29.4 m/s.

Alternative Method

This solution is always visible.

You can also solve this using energy conservation…

Nested Admonitions#

You can nest admonitions within each other:

Challenge Problem

Prove that \(\sqrt{2}\) is irrational.

Hint

Use proof by contradiction. Assume \(\sqrt{2} = \frac{p}{q}\) where \(p\) and \(q\) are coprime integers.

Løsning

Proof by contradiction:

Assume \(\sqrt{2} = \frac{p}{q}\) where \(p\) and \(q\) are coprime…

[Rest of proof here]

Conclusion: This contradicts our assumption, so \(\sqrt{2}\) is irrational.

Features

All these admonitions support:

  • LaTeX math: Both inline \(x^2\) and display $\(\int_0^1 x dx\)$

  • Markdown: Including bold, italic, and code

  • Dropdown: Optional collapsible sections

  • Theme awareness: Adapts to light/dark mode with custom icons

  • Custom titles: Most directives allow custom titles

  • MyST syntax: Full support for colon-fence syntax (:::)

Contents