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CIE A-Level Maths Study Notes

2.6.1 Root Approximation Techniques

Root Approximation Techniques form an integral part of Mathematics, providing essential tools for solving equations. These techniques are widely applied in various fields, from engineering to economics. This section delves into the different methods used to approximate roots, including graphical, numerical, and advanced iterative methods.

Graphical Methods for Identifying Roots

1. Plot the Function: Graph the function to visualise where it intersects the x-axis. For instance, plot f(x)=x36x2+11x6f(x) = x^3 - 6x^2 + 11x - 6.

2. Identify Intersections: The x-axis intersections indicate the approximate roots of the equation.

3. Refine the Approximation: Zoom in on these intersections for a more precise approximation.

4. Use a Table of Values: Confirm the roots by examining sign changes in a table of values.

Example: Graphical Root Finding

Problem: Approximate the roots of x36x2+11x6=0x^3 - 6x^2 + 11x - 6 = 0 graphically.

Solution:

  • Plotting the function and observing intersections provides approximate root locations.
  • Zoom in for accuracy and confirm with a table of values, checking for sign changes.
Plot of the Function

To confirm the roots of the function f(x)=x36x2+11x6f(x) = x^3 - 6x^2 + 11x - 6, we can use a table of values.

table of values

From this table, we can observe the following:

  • At x=1.0x = 1.0, f(x)f(x) changes from negative to positive, indicating a root at x=1x = 1.
  • At x=2.0x = 2.0, f(x)f(x) is exactly 0, confirming a root at x=2.x = 2.
  • At x=3.0x = 3.0, f(x)f(x) changes from negative to positive, indicating a root at x=3x = 3.

Numerical Investigation of Sign Changes

Process of Numerical Investigation

1. Select a Range: Choose a suspected range for the roots, e.g., from x=1x = 1 to x=4x = 4.

2. Compute Values: Calculate f(x)f(x) for each point in the range.

3. Detect Sign Changes: Look for intervals where f(x)f(x) changes sign, indicating a root.

Example: Numerical Sign Change

Problem: Find an interval containing a root for f(x)=x25x+6f(x) = x^2 - 5x + 6.

Solution:

Step 1: Select a Range

  • Range: x=1x = 1 to x=4x = 4

Step 2: Compute Values

  • For x=1:f(1)=125×1+6=15+6=2x = 1 : f(1) = 1^2 - 5 \times 1 + 6 = 1 - 5 + 6 = 2
  • For x=2:f(2)=225×2+6=410+6=0x = 2 : f(2) = 2^2 - 5 \times 2 + 6 = 4 - 10 + 6 = 0
  • For x=3:f(3)=325×3+6=915+6=0x = 3 : f(3) = 3^2 - 5 \times 3 + 6 = 9 - 15 + 6 = 0
  • For x=4:f(4)=425×4+6=1620+6=2x = 4 : f(4) = 4^2 - 5 \times 4 + 6 = 16 - 20 + 6 = 2

Step 3: Detect Sign Changes

  • Between x=1x = 1 and x=2x = 2, f(x)f(x)changes from positive 22 to zero.
  • Between x=2x = 2 and x=3x = 3, f(x)f(x) remains zero.
  • Between x=3x = 3 and x=4x = 4, f(x)f(x) remains positive 2.2.

Practical Examples: Bracketing a Root

Steps for Bracketing a Root

1. Choose Integers: Select a range of consecutive integers for evaluation.

2. Evaluate the Function: Compute f(x)f(x)at each integer.

3. Find the Bracket: Identify intervals where f(x)f(x)'s sign changes, indicating a root.

Example: Root Bracketing

Problem: Bracket a root for x32x25x+6=0x^3 - 2x^2 - 5x + 6 = 0.

Solution:

Step 1: Choose Integers

  • Selected integers: 2,1,0,1,2-2, -1, 0, 1, 2

Step 2: Evaluate the Function

  • For x=2:f(2)=(2)32(2)25(2)+6=8+8+10+6=16x = -2 : f(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 = -8 + 8 + 10 + 6 = 16
  • For x=1:f(1)=(1)32(1)25(1)+6=12+5+6=8x = -1 : f(-1) = (-1)^3 - 2(-1)^2 - 5(-1) + 6 = -1 - 2 + 5 + 6 = 8
  • For x=0:f(0)=032×025×0+6=6 x = 0 : f(0) = 0^3 - 2 \times 0^2 - 5 \times 0 + 6 = 6
  • For x=1:f(1)=132×125×1+6=0 x = 1 : f(1) = 1^3 - 2 \times 1^2 - 5 \times 1 + 6 = 0
  • For x=2:f(2)=232×225×2+6=8810+6=4x = 2 : f(2) = 2^3 - 2 \times 2^2 - 5 \times 2 + 6 = 8 - 8 - 10 + 6 = -4

Step 3: Find the Bracket

  • No sign change is observed between x=2x = -2 and x=1x = -1 since f(2)=16f(-2) = 16 and f(1)=8f(-1) = 8, both positive.
  • No sign change is observed between x=1x = -1 and x=0x = 0 since both f(1)f(-1) and f(0)f(0)are positive.
  • A sign change is observed between x=0x = 0 and x=1x = 1 since f(0)=6f(0) = 6(positive) and f(1)=0f(1) = 0 (zero).
  • A sign change is observed between x=1x = 1 and x=2x = 2 since f(1)=0f(1) = 0(zero) and f(2)=4f(2) = -4 (negative), indicating a root between x=1x = 1 and x=2x = 2.
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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