Change of signs method for SQA Higher Maths
This page covers the following topics:
1. Graphical representation of the change of signs method
2. Change of signs method
3. Limits of the change of sign method
4. Change of signs method problems in context
Suppose we have 2 points a and b, and some function f(x). Provided f(x) is a continuous function, if f(a) < 0 and f(b) > 0, then we know f(x) crosses the x–axis between a and b, therefore having a root in that interval. This is increadibly obvious if one looks at a graph of f(x).
Consider the graph shown with f(a) < 1 and f(b) > 1. Any continuous line f(x) between a and b must cross the point f(x) = 1. The change of sign method can be used to locate solutions for any equation, provided it can be rearranged to the form f(x) = 0. Also, using the change of sign method can help solve an equation. One can just strategically guess roots and narrow down the interval we know the root is in. In general, for more complicated functions, the change of sign method can be really useful to find where a root is, without analytically solving the equation for f(x) = 0.
There's some issues with this method one should be aware of. Firstly, there are some occasions a function has a root without a change of signs. Consider the function shown, which never has a change of sign, but still has a root. A change of sign means there is a root – but having a root does not mean there has to be a change of sign. If your interval is big enough, the function may have a couple of roots between 2 points where f(x) has the same sign. Also, the change of sign method only works with continuous functions – there can't be any jumps or gaps.
The change of signs method can also be used in more problem solving contexts. It can be useful for any continuous function which could, for example, give a velocity value. Perhaps one would use the change of signs method on an equation which was the result of equating 2 equations together (for the intersection of them both).
Give a root of f(x) = x² + 8x + 12. Instead of factorising, make use of the change of signs method.
Trying a couple of values, f(0) = 12 > 0, f(–3) = –3 < 0. We now know an interval the root is in. Trying the value x = 2 gives a root.
Prove x + cos(x) = sin(x) has a solution between x = –2 and x = –1.
First rearrange to f(x) = x + cos(x) – sin(x). Then show there is a change of sign between f(–2) < 0 and f(–1) > 0.
Give a root of f(x) = x⁴ + x³ – 8, making use of the change of signs method.
One can just try a bunch of values. One can find f(–3) > 0, and f(0) < 0. Here there is a change of signs so we know there is a root in this interval. One can just guess f(–2), which equals 0. So x = 2 is a root.
Show that x³ + cos(x) = 0 has a solution between x = –1 and x = 0.
Consider the left hand side of the equation as f(x). We have f(–1) < 0 and f(0) > 0. There is a change of signs so the left hand side must equal zero at some point in this range.
Prove x⁴ – sin(x) = 3 must have a solution between x = 1 and x = 2.
First rearrange to f(x) = x⁴ – sin(x) – 3. Then show f(x) has a change of sign between f(1) < 0 and f(2) > 0.
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