# Tangents, normals, critical points, inflection points

> IB Mathematics AI HL · IB AI HL Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-tangents-normals-critical-points-inflection/

This sub-topic teaches core applications of differentiation: finding tangent and normal equations, and locating critical points and inflection points. These are high-frequency exam skills that form the basis for optimization and curve sketching.

**Prerequisites:** [Basic differentiation of common functions](https://www.owlsprep.com/study/ib-math-ai-hl-u3-basic-differentiation/); [Calculating first and second derivatives](https://www.owlsprep.com/study/ib-math-ai-hl-u4-first-second-derivatives/)

## Learning objectives

- Find equations of tangents and normals to any differentiable curve
- Locate critical (stationary) points by solving $f'(x) = 0$
- Classify critical points as local maxima/minima using the second derivative test
- Identify and verify inflection points where concavity changes

## Tangents and Normals to Curves

**Tangent and Normal Lines** — The tangent to a curve at $x=a$ has gradient equal to the first derivative of the function at $x=a$. The normal is perpendicular to the tangent at this point, so its gradient is the negative reciprocal of the tangent gradient.

*Example:* If tangent gradient $m_t = 2$, normal gradient $m_n = -\frac{1}{2}$

**Worked example:** Find the equation of the tangent and normal to $y = x^2 - 3x + 2$ at $x=2$.

1. First find the $y$-coordinate of the point:
2. $$y = (2)^2 - 3(2) + 2 = 0$$
3. The point is $(2, 0)$. Find the gradient function:
4. $$\frac{dy}{dx} = 2x - 3$$
5. Evaluate gradient for the tangent at $x=2$:
6. $$m_t = 2(2) - 3 = 1$$
7. Gradient of the perpendicular normal is:
8. $$m_n = -\frac{1}{m_t} = -1$$
9. Use point-gradient form $y - y_1 = m(x - x_1)$ for both lines:
10. Tangent: $y = x - 2$  
Normal: $y = -x + 2$

> **tip**
>
> Always use the original function to get the $y$-coordinate, never the derivative. Derivatives give gradient, not function values.

> **Exam tip:** Always write your final line equation in the form $ax + by + c = 0$ if requested, you will lose marks for incorrect form.

*Calculator:* allowed

## Critical (Stationary) Points

**Critical (Stationary) Point** — A point on a curve where the first derivative equals zero: $f'(a) = 0$. This means the tangent at the point is horizontal (gradient = 0).

*Example:* All local maxima and minima are critical points.

**Worked example:** Find all critical points of $f(x) = x^3 - 6x^2 + 9x + 1$.

1. Calculate the first derivative:
2. $$f'(x) = 3x^2 - 12x + 9$$
3. Set $f'(x) = 0$ and solve for $x$:
4. $$3x^2 - 12x + 9 = 0 \implies 3(x-1)(x-3) = 0$$
5. Solutions are $x=1$ and $x=3$. Calculate corresponding $y$-values:
6. $f(1) = 1 - 6 + 9 + 1 = 5 \implies (1, 5)$  
$f(3) = 27 - 54 + 27 + 1 = 1 \implies (3, 1)$
7. The critical points are $(1, 5)$ and $(3, 1)$.

*Calculator:* allowed

## Classifying Critical Points

The most common method for classifying critical points in IB exams is the **second derivative test**: if $f'(a) = 0$, evaluate $f''(a)$ to determine the nature of the point:

- If $f''(a) < 0$: the point is a **local maximum** (concave down)
- If $f''(a) > 0$: the point is a **local minimum** (concave up)
- If $f''(a) = 0$: the test is inconclusive, use the first derivative sign test

**Worked example:** Classify the critical points $(1,5)$ and $(3,1)$ from the previous example using the second derivative test.

1. Find the second derivative from $f'(x) = 3x^2 - 12x + 9$:
2. $$f''(x) = 6x - 12$$
3. Evaluate $f''(x)$ at $x=1$:
4. $$f''(1) = 6(1) - 12 = -6 < 0$$
5. Since $f''(1) < 0$, $(1, 5)$ is a local maximum.
6. Evaluate $f''(x)$ at $x=3$:
7. $$f''(3) = 6(3) - 12 = 6 > 0$$
8. Since $f''(3) > 0$, $(3, 1)$ is a local minimum.

*Calculator:* allowed

## Inflection Points

**Inflection Point** — A point where the concavity of the curve changes. For an inflection point at $x=a$, $f''(a) = 0$ **and** the sign of $f''(x)$ changes around $x=a$.

*Example:* $f''(x)$ changes from negative to positive (or positive to negative) at the point.

**Worked example:** Find the inflection point of $f(x) = x^3 - 6x^2 + 9x + 1$.

1. We already know the second derivative:
2. $$f''(x) = 6x - 12$$
3. Set $f''(x) = 0$ and solve for $x$:
4. $$6x - 12 = 0 \implies x = 2$$
5. Check that the sign of $f''(x)$ changes around $x=2$:
6. For $x < 2$: $f''(x) < 0$ (concave down)  
For $x > 2$: $f''(x) > 0$ (concave up)
7. Concavity changes, so inflection point exists. Find its $y$-coordinate:
8. $$f(2) = 8 - 24 + 18 + 1 = 3$$
9. The inflection point is $(2, 3)$.

> **warning**
>
> $f''(a) = 0$ alone is not enough to confirm an inflection point. You must verify the sign change of $f''(x)$ to get full marks.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to calculate the $y$-coordinate of the point when finding tangent/normal equations
  - Why it fails: Point-gradient form requires both coordinates of the point to find the full equation
  - Correct: Always substitute $x=a$ into the original function to get $y=f(a)$ before finding the line equation
- **Wrong:** Using the tangent gradient for the normal, or taking reciprocal instead of negative reciprocal
  - Why it fails: Perpendicular lines have gradients that multiply to $-1$, not $1$
  - Correct: If tangent gradient is $m_t$, normal gradient is $m_n = -\frac{1}{m_t}$ (for $m_t \neq 0$)
- **Wrong:** Assuming all critical points are either maxima or minima
  - Why it fails: Horizontal points of inflection also have $f'(a) = 0$ but are not turning points
  - Correct: Always classify all critical points using the first or second derivative test
- **Wrong:** Claiming an inflection point exists just because $f''(a) = 0$, with no sign check
  - Why it fails: For $f(x) = x^4$, $f''(0) = 0$ but concavity does not change, so there is no inflection point
  - Correct: Always test the sign of $f''(x)$ on both sides of $x=a$ to confirm a concavity change

## Cheatsheet

| Concept | Key Condition | Result/Equation |
| --- | --- | --- |
| Tangent at $x=a$ | $m_t = f'(a)$ | $y - f(a) = m_t(x-a)$ |
| Normal at $x=a$ | $m_n = -\frac{1}{f'(a)}, f'(a) \neq 0$ | $y - f(a) = m_n(x-a)$ |
| Critical point | $f'(a) = 0$ | Horizontal tangent at $(a, f(a))$ |
| Local maximum | $f'(a) = 0, f''(a) < 0$ | Concave down turning point |
| Local minimum | $f'(a) = 0, f''(a) > 0$ | Concave up turning point |
| Inflection point | $f''(a) = 0$, concavity changes | Change in curve curvature direction |

## What's next

The skills covered in this subtopic are the foundation for all applied differentiation questions in IB AI HL. You will use tangents and normals to solve problems about rates of change, critical points to solve practical optimization problems, and all these concepts to sketch accurate curves from derivative information. Mastery of this subtopic is essential to score full marks on extended response calculus questions, which make up a large portion of exam marks.

- [Optimization Problems](https://www.owlsprep.com/study/ib-math-ai-hl-u4-optimization-problems/)
- [Related Rates of Change](https://www.owlsprep.com/study/ib-math-ai-hl-u4-related-rates/)
- [Monotonic functions, second derivative test](https://www.owlsprep.com/study/ib-math-ai-hl-u4-monotonic-functions-second-derivative-test/)

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