Tangents, normals, critical points, inflection points

IB Mathematics AI HL· 4.4 Applications of Differentiation· 15 min read

1. Tangents and Normals to Curves★★☆☆☆⏱ 5 min

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📘 Definition

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 $y = x^{2} - 3 x + 2$ 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: $y = - x + 2$

Exam tip:

Always write your final line equation in the form $a x + b y + c = 0$ if requested, you will lose marks for incorrect form.

2. Critical (Stationary) Points★★☆☆☆⏱ 4 min

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📘 Definition

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 $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

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
$f (1) = 1 - 6 + 9 + 1 = 5 ⟹ (1, 5)$
$f (3) = 27 - 54 + 27 + 1 = 1 ⟹ (3, 1)$
7
The critical points are $(1, 5)$ and $(3, 1)$ .

3. Classifying Critical Points★★★☆☆⏱ 6 min

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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 $f^{'} (x) = 3 x^{2} - 12 x + 9$ :
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.

4. Inflection Points★★★☆☆⏱ 5 min

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📘 Definition

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 $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

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 $(2, 3)$ .

5. Common Pitfalls

Wrong move:

Forgetting to calculate the $y$ -coordinate of the point when finding tangent/normal equations

Why:

Point-gradient form requires both coordinates of the point to find the full equation

Correct move:

Always substitute $x = a$ into the original function to get $y = f (a)$ before finding the line equation

Wrong move:

Using the tangent gradient for the normal, or taking reciprocal instead of negative reciprocal

Why:

Perpendicular lines have gradients that multiply to $- 1$ , not $1$

Correct move:

If tangent gradient is $m_{t}$ , normal gradient is $m_{n} = - \frac{1}{m _{t}}$ (for $m_{t} \neq = 0$ )

Wrong move:

Assuming all critical points are either maxima or minima

Why:

Horizontal points of inflection also have $f^{'} (a) = 0$ but are not turning points

Correct move:

Always classify all critical points using the first or second derivative test

Wrong move:

Claiming an inflection point exists just because $f^{''} (a) = 0$ , with no sign check

Why:

For $f (x) = x^{4}$ , $f^{''} (0) = 0$ but concavity does not change, so there is no inflection point

Correct move:

Always test the sign of $f^{''} (x)$ on both sides of $x = a$ to confirm a concavity change

6. Quick Reference 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Find tangent to cubic curve at given point
2024 · 2
Classify critical points of a quartic function

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.