Linear Interpolation (Edexcel International A Level Further Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

Linear Interpolation

What is linear interpolation?

Linear interpolation draws a straight line between the points A and B with coordinates $(a, f (a))$ and $(b, f (b))$ , where $a < x < b$ is an interval containing a root to the equation $f (x) = 0$
- As $f (a)$ and $f (b)$ have different signs, A and B are on either side of the x-axis
- The approximation to the root is where this straight line cuts the x-axis
  - This is at the point X with coordinates $(x, 0)$
  - Points A, X and B lie on the same straight line
It helps to use the gradient formula
- The gradient between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

How do I apply linear interpolation using gradients?

This is best explained with an example
If the root of $f (x) = 0$ lies in the interval $[1, 2]$ and $f (1) = - 5$ and $f (2) = 10$ , then:
- A is the point $(1, - 5)$
- B is the point $(2, 10)$
- X is the point $(x, 0)$
  - This is the x-intercept of the line AB
  - A, X and B lie on the same straight line
- The gradient of AX equals the gradient of XB
  - The order matters (not BX)
- Use the gradient formula to set these equal
  - $\frac{0 - (- 5)}{x - 1} = \frac{10 - 0}{2 - x}$ giving $\frac{5}{x - 1} = \frac{10}{2 - x}$
- Solve the equation to find $x$
  - $5 (2 - x) = 10 (x - 1)$
  - expand and solve to get $x = \frac{4}{3}$
This process can be repeated to get the next approximation
This gradient method does not require a sketch

What other methods can I use to do linear interpolation?

You can sketch the points A, B and X on a diagram and use similar triangles either side of the x-axis to find an equation in $x$
- This leads to the same equation as the gradients above
Or you can use coordinate geometry to find the equation of the line AB
- Use $y - y_{1} = m (x - x_{1})$
- Then find its x-intercept
  - Substitute in $y = 0$ and solve for $x$

Exam Tip

If the calculations involve lengthy decimals, you can use $f (a)$ and $f (b)$ to stand for their numbers in your working!

Worked Example

The equation $f (x) = 0$ where $f (x) = x^{3} - \sqrt{x} - 2$ has a root in the interval $[1.4, 1.5]$ .

Use linear interpolation once to find an approximation to the root, giving your answer to 3 decimal places.

Find $f (1.4)$ and $f (1.5)$

$\begin{array}{rcl} f (1.4) & = & 1 . 4^{3} - \sqrt{1.4} - 2 = - 0.439215 . . . \\ f (1.5) & = & 1 . 5^{3} - \sqrt{1.5} - 2 = 0.150255 . . . \end{array}$

It is easier to write them as $f (1.4)$ and $f (1.5)$ for now
Let A, B and X be the points $(1.4, f (1.4))$ , $(1.5, f (1.5))$ and $(x, 0)$
Use the gradient formula $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ to find the gradient of AX

gradient of AX is $\frac{0 - f (1.4)}{x - 1.4}$

Use the gradient formula $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ to find the gradient of XB

gradient of XB is $\frac{f (1.5) - 0}{1.5 - x}$

Set the two gradients equal to each other

$\frac{0 - f (1.4)}{x - 1.4} = \frac{f (1.5) - 0}{1.5 - x}$

Simplify and cross multiply

$\begin{array}{rcl} \frac{- f (1.4)}{x - 1.4} & = & \frac{f (1.5)}{1.5 - x} \\ - f (1.4) (1.5 - x) & = & f (1.5) (x - 1.4) \end{array}$

Expand and collect the $x$ terms on the right-hand side

$\begin{array}{rcl} - 1.5 f (1.4) + f (1.4) x & = & f (1.5) x - 1.4 f (1.5) \\ 1.4 f (1.5) - 1.5 f (1.4) & = & f (1.5) x - f (1.4) x \end{array}$

Factorise out the $x$ , then make $x$ the subject

$\begin{array}{rcl} 1.4 f (1.5) - 1.5 f (1.4) & = & x [f (1.5) - f (1.4)] \\ \frac{1.4 f (1.5) - 1.5 f (1.4)}{f (1.5) - f (1.4)} & = & x \end{array}$

Substitute in $f (1.4)$ and $f (1.5)$ from above

$\begin{array}{rcl} \frac{1.4 \times 0.150255 . . . - 1.5 \times (- 0.439215 . . .)}{0.150255 . . . - (- 0.439215 . . .)} & = & x \\ 1.47451 . . . & = & x \end{array}$

Give your final answer to 3 decimal places

$1.475$ to 3 decimal places

If you did the working without $f (1.4)$ and $f (1.5)$ , keep lots of decimal places

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

Downloadable PDFs
Unlimited Revision Notes
Topic Questions
Past Papers
Model Answers
Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Next topic

Did this page help you?