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Year 11 Maths

Curve Sketching with Calculus

Use derivatives to find stationary points, determine their nature, and sketch accurate graphs of functions.

Finding Stationary Points

A stationary point occurs where the gradient of the curve is zero, that is, where f'(x) = 0. At these points, the tangent line is horizontal.

Steps to Find Stationary Points

1

Find the first derivative f'(x).

2

Set f'(x) = 0 and solve for x.

3

Find the y-coordinates by substituting x-values back into f(x).

Determining the Nature of Stationary Points

There are two methods to determine whether a stationary point is a local maximum, local minimum, or point of inflection:

Method 1: Second Derivative Test

  • If f''(x) < 0 at the stationary point: local maximum
  • If f''(x) > 0 at the stationary point: local minimum
  • If f''(x) = 0: test is inconclusive -- use Method 2

Method 2: Sign Diagram of f'(x)

  • If f'(x) changes from + to -: local maximum
  • If f'(x) changes from - to +: local minimum
  • If f'(x) does not change sign: horizontal point of inflection

Sign Diagram Example

For f(x) = x3 - 3x: f'(x) = 3x2 - 3 = 3(x - 1)(x + 1)

+ + +

x < -1

increasing

0

x = -1

max

- - -

-1 < x < 1

decreasing

0

x = 1

min

+ + +

x > 1

increasing

Curve Sketching Checklist

To produce a thorough curve sketch, follow these steps systematically:

1

Find the y-intercept: set x = 0 and find f(0).

2

Find the x-intercepts: set f(x) = 0 and solve.

3

Find stationary points: solve f'(x) = 0, then determine their nature.

4

Consider end behaviour: what happens as x approaches positive and negative infinity?

5

Plot key points and connect with a smooth curve showing correct concavity.

Key Vocabulary

Stationary Point

A point on a curve where f'(x) = 0. The tangent is horizontal. Can be a local max, local min, or point of inflection.

Local Maximum

A turning point where the function changes from increasing to decreasing. f'(x) changes from positive to negative.

Local Minimum

A turning point where the function changes from decreasing to increasing. f'(x) changes from negative to positive.

Point of Inflection

A point where the concavity changes. At a horizontal point of inflection, f'(x) = 0 but does not change sign.

Worked Examples

1

Find and classify the stationary points of f(x) = x3 - 12x + 2.

Step 1: f'(x) = 3x2 - 12. Set f'(x) = 0: 3x2 - 12 = 0, so x2 = 4, giving x = 2 or x = -2.

Step 2: f''(x) = 6x. At x = 2: f''(2) = 12 > 0, so local minimum. At x = -2: f''(-2) = -12 < 0, so local maximum.

Step 3: f(2) = 8 - 24 + 2 = -14. f(-2) = -8 + 24 + 2 = 18.

Answer: Local maximum at (-2, 18); local minimum at (2, -14).

2

Sketch y = x3 - 3x2 showing all key features.

y-intercept: f(0) = 0. So the curve passes through (0, 0).

x-intercepts: x2(x - 3) = 0, so x = 0 (double root) or x = 3.

Stationary points: f'(x) = 3x2 - 6x = 3x(x - 2) = 0 at x = 0 and x = 2.

Nature: f''(x) = 6x - 6. f''(0) = -6 < 0 (max at (0, 0)). f''(2) = 6 > 0 (min at (2, -4)).

Answer: Local max at (0, 0), local min at (2, -4), x-intercepts at 0 and 3.

3

Show that y = x3 has a horizontal point of inflection at x = 0.

Step 1: f'(x) = 3x2. At x = 0: f'(0) = 0 (stationary point).

Step 2: f''(x) = 6x. At x = 0: f''(0) = 0 (inconclusive).

Step 3: Sign diagram of f'(x) = 3x2: f'(x) ≥ 0 for all x. It does not change sign at x = 0.

Answer: Since f'(x) does not change sign, (0, 0) is a horizontal point of inflection.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

Where are the stationary points of f(x) = x2 - 6x + 5?

Question 2

If f''(a) < 0 at a stationary point x = a, the point is:

Question 3

For f(x) = 2x3 + 3x2 - 12x + 1, find the x-coordinates of the stationary points.

Question 4

If f'(x) changes from negative to positive at x = a, then x = a is:

Question 5

What is the y-intercept of f(x) = x3 - 4x + 7?

Key Concepts Summary

Rates of Change Optimisation Problems