Intermediate Value Theorem

MathJax example

Once we know that a root lies within a particular interval we need a process to help us to determine the value of the root or an approximation to the root. To do this we use ITERATIVE processes. An ITERATIVE process is a repeated calculation which allows us get closer to the desired result with each iteration.

If you had to guess a value for one of the roots of the equation LaTeX: 2x^3-5x^2+x+1=0 what would you guess? That's quite a challenge. It may be easier to try to guess a range of values in which you think the root occurs.

That's where the Intermediate Value Theorem comes in. This Theorem allows us to determine if a root occurs within a specified range.

THE INTERMEDIATE VALUE THOREM

If $LaTeX: f\left(x\right)$ is a continuous function on the closed interval $LaTeX: \left[a,\:b\right]$ and the product $LaTeX: f\left(a\right)f\left(b\right)<0$ then there LaTeX: c exists in $LaTeX: \left[a,\:b\right]$ such that $LaTeX: f\left(c\right)=0$

EXAMPLE 1

Use the Intermediate Value Theorem to show that LaTeX: f(x)=x^3-2x^2+x-1 has a root between 1 and 2.

SOLUTION

is a polynomial and therefore continuous on the interval LaTeX: [1,2]

.
$LaTeX: f(1)\times f(2)=-1$
By the Intermediate Value Theorem there must be some $LaTeX: c\in\left[1,\:2\right]$ such that LaTeX: f(c)=0

. Therefore, there is a root between 1 and 2.

VIDEO SOLUTION

EXAMPLE 2

Use the Intermediate Value Theorem to verify that there is a root of the equation $LaTeX: 3\sqrt{x}+\frac{18}{\sqrt{x}}=20$ between 1.1 and 1.2.

SOLUTION
Firstly we need to rewrite the equation so that we have LaTeX: f(x)=0 .
$LaTeX: 3\sqrt{x}+\frac{18}{\sqrt{x}}=20$
$LaTeX: 3\sqrt{x}+\frac{18}{\sqrt{x}}-20=0$
Therefore, $LaTeX: f\left(x\right)=3\sqrt{x}+\frac{18}{\sqrt{x}}-20$

$LaTeX: f\left(1.1\right)=3\sqrt{1.1}+\frac{18}{\sqrt{1.1}}-20=0.309$

$LaTeX: f\left(1.2\right)=3\sqrt{1.2}+\frac{18}{\sqrt{1.2}}-20=-0.281$
NB: LaTeX: f(x) is NOT a polynomial so we do not make that assertion in our statement for the INTERMEDIATE VALUE THEOREM.

LaTeX: f(x) is a continuous on the interval LaTeX: [1.1,1.2] .

LaTeX: f(1.1)f(1.2)<0
By the Intermediate Value Theorem there must be some $LaTeX: c\in[1.1,1.2]$ such that LaTeX: f(c)=0 . Therefore, there is a root between 1.1 and 1.2.