Showing a Solution Exists in an Interval

Grade 6

Why we need to show a root exists

Before iterating, exam questions often ask you to justify that a solution lies in a given interval, say between x = 1 and x = 2. The method is the change of sign test: if f(x) is continuous and changes sign between two values, it must cross zero between them.

The method

Evaluate f(x) at both endpoints. If one is positive and the other is negative, the function has crossed the x-axis — meaning there is a root in that interval.

You must state the values of f at each endpoint and explicitly say "change of sign, therefore a root exists in the interval." Just saying "one is positive, one is negative" without the concluding sentence will not earn the mark.

Worked example

EXAMPLE

Show that x³ − 2x − 5 = 0 has a root between x = 2 and x = 3.

Let f(x) = x³ − 2x − 5. f(2) = 8 − 4 − 5 = −1 (negative). f(3) = 27 − 6 − 5 = 16 (positive). There is a change of sign, and f is continuous, so there is a root between x = 2 and x = 3.

Confirming a root to a given accuracy

To confirm that x = 2.79 is correct to 2 decimal places, test f at the bounds of the rounding interval: evaluate f(2.785) and f(2.795). If there is a change of sign between them, the root is between 2.785 and 2.795, and rounds to 2.79. This is the standard "trap it" method examiners expect.

Try this one

Show that the equation \( x^3 - 3x + 1 = 0 \) has a root between \( x = 1 \) and \( x = 2 \). [2]

Nova's hint:

This is a change of sign method question — you need to show the equation changes from positive to negative (or vice versa) in the given interval.

Ask yourself: 'What happens when I substitute both \( x = 1 \) and \( x = 2 \) into the expression — do I get one positive and one negative value?'

For example, to show a root between \( x = 0 \) and \( x = 1 \) for \( f(x) = x^2 - x - 0.5 \), calculate \( f(0) \) and \( f(1) \) and check the signs differ.

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