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Locating roots

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Explainer Video

Tutor: Daniel

Summary

Locating roots

In a nutshell

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



Sign changes

Consider the function f(x)=4x3−10x2+2f(x)=4x^3-10x^2+2f(x)=4x3−10x2+2 and its graph below.

Maths; Numerical methods; KS5 Year 13; Locating roots

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


Example 1

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

​f(x)=4x3−10x2+2f(x)=4x^3-10x^2+2f(x)=4x3−10x2+2

​​

x=−1f(−1)=4(−1)3−10(−1)2+2f(−1)=−12‾\begin{aligned}x&=-1\\f(-1)&=4(-1)^3-10(-1)^2+2\\f(-1)&=\underline{-12}\end{aligned}xf(−1)f(−1)​=−1=4(−1)3−10(−1)2+2=−12​​​​​

​x=0f(0)=4(0)3−10(0)2+2f(0)=2‾\begin{aligned}x&=0\\f(0)&=4(0)^3-10(0)^2+2\\f(0)&=\underline{2}\\\end{aligned}xf(0)f(0)​=0=4(0)3−10(0)2+2=2​​​​


Since f(x)f(x)f(x)​ undergoes a sign change from negative to positive (from −12-12−12 to 222) between the interval [−1,0][-1,0][−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][-1,1][−1,1]. As you already know the value of f(−1)f(-1)f(−1), let's find f(1)f(1)f(1).


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


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


Example 2

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


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


There is a sign change between f(−1)f(-1)f(−1) and f(3)f(3)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)=13−xf(x)=\dfrac{1}{3-x}f(x)=3−x1​, where when x>3x>3x>3 will give a negative f(x)f(x)f(x) and when x<3x < 3x<3 will give a positive f(x)f(x)f(x), but x=3x=3x=3 is undefined and is not a root. There is not a root in the interval [2,4][2,4][2,4] even though there is a change in sign.

Maths; Numerical methods; KS5 Year 13; Locating roots


Example 4

A function f(x)=−3x3+8x2−3f(x)=-3x^3+8x^2-3f(x)=−3x3+8x2−3 has three roots.

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

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


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


​f(x)=−3x3+8x2−3f(x)=-3x^3+8x^2-3f(x)=−3x3+8x2−3

x=0f(0)=−3(0)3+8(0)2−3f(0)=−3\begin{aligned}x&=0\\f(0)&=-3(0)^3+8(0)^2-3\\f(0)&=-3\\\end{aligned}xf(0)f(0)​=0=−3(0)3+8(0)2−3=−3​​​

​x=3f(3)=−3(3)3+8(3)2−3f(3)=−12\begin{aligned}x&=3\\f(3)&=-3(3)^3+8(3)^2-3\\f(3)&=-12\\\end{aligned}xf(3)f(3)​=3=−3(3)3+8(3)2−3=−12​​​


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


(ii) To prove that α=2.508\alpha=2.508α=2.508​, show that there is a sign change in the interval [2.5075,2.5085][2.5075,2.5085][2.5075,2.5085], as the numbers between this interval round to 2.5082.5082.508.


f(x)=−3x3+8x2−3f(2.5075)≈0.00231f(2.5085)≈−0.0142\begin{aligned}f(x)&=-3x^3+8x^2-3\\f(2.5075)&\approx0.00231\\f(2.5085)&\approx-0.0142\end{aligned}f(x)f(2.5075)f(2.5085)​=−3x3+8x2−3≈0.00231≈−0.0142​


As there is a sign change between f(2.5075)f(2.5075)f(2.5075) and f(2.5085)f(2.5085)f(2.5085), you can say that 2.5075<α<2.50852.5075<\alpha <2.50852.5075<α<2.5085. Therefore α=2.508\alpha=2.508α=2.508 when rounded to three decimal places.


​2.5075<α<2.5085 ⟹ α≈2.508‾\underline{2.5075<\alpha <2.5085 \implies \alpha \approx 2.508}2.5075<α<2.5085⟹α≈2.508​

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FAQs - Frequently Asked Questions

Does a function with an asymptote have roots?

A function with an asymptote may have roots, but not at the asymptote. The f(x) can change signs at either side of the asymptote, however.

How do you find a function's roots?

You can use intervals to locate the approximate location of a function's roots. A sign change between f(x) values over an interval will imply an odd number of roots between the interval. No sign change between f(x) values implies an even number of roots, or no roots at all.

What is a function's root?

A function's root is the x value where f(x)=0, or in other words where the function crosses the x-axis.

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