Locating roots

In a nutshell

A function's root is the $$x$$ value where $$f(x)=0$$, or in other words where the function crosses the $$x$$-axis. You can use intervals to locate the approximate location of a function's roots. You will use sign changes in $$f(x)$$ values to indicate how many roots a function has and where those roots are.

Sign changes

Consider the function $$f(x)=4x^3-10x^2+2$$ and its graph below.

This function intersects the $$x$$-axis three times, meaning it has three roots. You can use intervals to show mathematically that a function has roots. Observe the following rule:

If a continuous function $$\bf f(x)$$ is examined between the interval $$\bf [n,m]$$, such that $$\bf f(n)$$ and $$\bf f(m)$$ have different signs (one being negative and the other positive), then at least one root lies in the interval $$\bf [n,m].$$

Example 1

Between the interval $$[-1,0]$$, the function crosses the $$x$$-axis. Find the $$f(x)$$ value for this function when $$x=-1$$ and $$x=0$$.

​$$f(x)=4x^3-10x^2+2$$

​​

Since $$f(x)$$​ undergoes a sign change from negative to positive (from $$-12$$ to $$2$$) between the interval $$[-1,0]$$, you know there is at least one root there.

Failures of sign changes

It's important to note that a sign change gives you incomplete information. Consider the interval $$[-1,1]$$. As you already know the value of $$f(-1)$$, let's find $$f(1)$$.

$$\begin{aligned}f(1)&=4(1)^3 -10(1)^2 +2\\f(1)&=4 -10 +2\\f(1)&=-4\\\end{aligned}$$

There's no sign change between $$f(-1)$$ and $$f(1)$$, however you can see two roots in the interval $$[-1,1]$$. An even number of roots between an interval will show no change in sign. 

Example 2

Consider the interval $$[-1,3]$$. Find $$f(3)$$.

​$$\begin{aligned}f(3)&=4(3)^3 -10(3)^2 +2\\f(3)&=108 -90 +2\\f(3)&=\underline{20}\\\end{aligned}$$​​

There is a sign change between $$f(-1)$$ and $$f(3)$$. You can't conclude that there is only one root however: the graph shows the function actually has three roots in this interval. An odd number of roots between an interval will show a change in sign, but you will not know how many roots there are.

The final example of a failure in this method is an asymptote. Functions with asymptotes will have a change in sign, but will not have any roots. 

Example 3

Consider the function $$f(x)=\dfrac{1}{3-x}$$, where when $$x>3$$ will give a negative $$f(x)$$ and when $$x < 3$$ will give a positive $$f(x)$$, but $$x=3$$ is undefined and is not a root. There is not a root in the interval $$[2,4]$$ even though there is a change in sign.

Example 4

A function $$f(x)=-3x^3+8x^2-3$$ has three roots.

(i) Assuming there are roots, prove that an even number of roots exist in the interval $$[0,3]$$.

(ii) One of the roots of this function is $$\alpha$$. Prove that $$\alpha=2.508$$ when rounded to three decimal places.

(i) Find $$f(0)$$ and $$f(3)$$ to find the number of roots.

​$$f(x)=-3x^3+8x^2-3$$

There is no sign change between $$f(0)$$​ and $$f(3)$$, therefore there are an even number of roots.

(ii) To prove that $$\alpha=2.508$$​, show that there is a sign change in the interval $$[2.5075,2.5085]$$, as the numbers between this interval round to $$2.508$$.

$$\begin{aligned}f(x)&=-3x^3+8x^2-3\\f(2.5075)&\approx0.00231\\f(2.5085)&\approx-0.0142\end{aligned}$$

As there is a sign change between $$f(2.5075)$$ and $$f(2.5085)$$, you can say that $$2.5075<\alpha <2.5085$$. Therefore $$\alpha=2.508$$ when rounded to three decimal places.

​$$\underline{2.5075<\alpha <2.5085 \implies \alpha \approx 2.508}$$