The Intermediate Value Theorem: Definition, Formula, Examples

The Intermediate Value Theorem (IMVT) is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which significant insights of continuous functions are determined. One of the uses of the IVT is in finding the roots of equations. This concept of The Intermediate Value Theorem (IMVT) has been broadly applied in mathematics, physics, engineering, economics, and biology branches.

In this article, we will cover the concepts of the Intermediate Value Theorem. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023).

The Intermediate Value Theorem (IMVT)

Continuous Function: A real function $f$ is said to be continuous if it is continuous at every point in the domain of $f$.

This definition requires a bit of elaboration. Suppose $f$ is a function defined on a closed interval [a, b], then for $f$ to be continuous, it needs to be continuous at every point in $[\mathrm{a},\mathrm{b}]$ including the endpoints a and b .
Continuity of f at a means $\underset{x\to a^{+}}{lim}f(x)=f(a)$ and continuity of f at b $\underset{x\to b^{-}}{lim}f(x)=f(b)$

Observe that $\underset{x\to a^{-}}{lim}f(x)$ and $\underset{x\to b^{+}}{lim}f(x)$ do not make sense. As a consequence of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of $f$ is a singleton, $f$ is a continuous function.

The Intermediate Value Theorem is:
Let $f$ be a continuous function on the closed interval [a,b]. If $L$ is any value between $f(a)$ and $f(b)$, then there exists a number $c$ in the interval $(a,b)$ such that $f(c)=L$.

Let $f$ be continuous over a closed interval $[a,b]$ and $f(a)\ne f(b)$. If $z$ is any real number between $f(a)$ and $f(b)$, then there is at least one $x$ in $[a,b]$ satisfying $f(x)=z$

For example: Imagine you are walking along a path from point A to point $B$. If point $A$ is at a height of 100 meters and point $B$ is at a height of 200 meters, and if you move continuously without jumping, then at some point along your path, you must be at every height between 100 and 200 meters. The IVT essentially says that a continuous function must pass through every intermediate value between $f(a)$ and $f(b)$.

An important result from IMVT

If $f(x)$ is a continuous function in [a, b] and $f(a)$ and $f(b)$ are of opposite signs, then there is at least one root of $f(x)$ lying in (a, b).

Recommended Video Based on Intermediate Theorem

Solved Examples based on Intermediate Theorem

Example 1: Let $f(x)$ be a polynomial of degree 3 with real coefficients such that $f(-1)=-5,f(1)=3,f(2)=-2,f(4)=1$, then the number of real roots of $f(x)=0$ will be
1) 3
2) 1
3) 5
4) 2

Solution:
$\because f(x)$ is a polynomial, so it will be continuous everywhere.
Now $f(-1)$ and $f(1)$ are of the opposite sign so at least one root will lie in $(-1,1)$
Similarly at least one in $(1,2)$ and at least one in $(2,4)$
hence, at least three roots will be there
But as $f(x)$ is a 3rd-degree polynomial, it can't have more than 3 roots
So, exactly three real roots will be there for $f(x)$.
Hence, the answer is the option 1.

Example 2: Let $f(x)$ is a function which is continuous and defined in $[-4,5]$ such that $f(-4)=2,f(-2)=-3,f(0)=1,f(3)=-4$ and $f(5)=3$ then the equation $f(x)=0$ has at least
1) 3 roots
2) 4roots
3) 5 roots
4) 6roots

Solution:
$\because \mathrm{f}(-4)$ and $\mathrm{f}(-2)$ are of opposite sign so there will be at least one root in $(-4,-2)$.
Similarly at least one root in $(-2,0)$ least one root in $(0,3)$ and at least one root in $(3,5)$
so total at least 4 roots will be there

Example 3: $f(x)$ is continuous function defined in $[1,5]$ such that $f(2)=1$ and $f(4)=10$ then the number of solution of equation $f(x)=3$ in $(2,-4)$ will be
1) 1
2) at least 1
3) more than 1
4) 2

Solution: $f(2)=1$ and $f(4)=10$ and $f(x)$ is continuous, so $f(x)$ will take all values between 1 and 10 at least one in the interval $(2,4)$, so $f(x)$ will also be 3 , at least once in $(2,4)$

Example 4: $f(x)=(x-a)(x-b)-1=0($ Where $a<b)$ has
1) Both roots in $(-\mathrm{\infty },a)$
2) Both roots in $(a,b)$
3) Both roots in $(b,\mathrm{\infty })$
4) One in $(-\mathrm{\infty },a)$ and one in $(b,\mathrm{\infty })$

Solution:

As we learned in

Number of roots of a polynomial equation -
For a polynomial equation $P(x)=0$ if $P(a)$ and $P(b)$ are of opposite sign then an odd number of roots lie between $a$ and $b$, if they are of the same sign then either no root or even number of roots lie between them.

$\begin{array}{rl}& f(-\mathrm{\infty })=+\mathrm{\infty }\\ & f(a)=-1\\ & f(b)=-1\\ & f(\mathrm{\infty })=+\mathrm{\infty }\end{array}$

$\because f(-\mathrm{\infty })$ and $f(a)$ are of the opposite sign so at least one root lies in

$(-\mathrm{\infty },a)$
Similarly, at least one root lies in $(b,\mathrm{\infty })$
But, since it is a quadratic equation it can't have more than two roots so exactly one lies in $(-\mathrm{\infty },a)$ and exactly one lies in $(b,\mathrm{\infty })$

Hence, the answer is the option 4.

Example 5: Let $a_1<a_2<a_3<a_4$ then the number of real roots of equation $(x-a_1)(x-a_3)+(x-a_2)(x-a_4)=0$ equals
1) 0
2) 1
3) 2
4)$\mathrm{\infty }$

Solution:

As we learnt in

Number of roots of polynomial equation -
For a polynomial equation $P(x)=0$ if $P(a)$ and $P(b)$ are of opposite sign then odd number of roots lie between $a$ and $b$, if they are of same sign then either no root or even number of roots lie between them.

$\begin{array}{rl}& \text{ Let }f(x)=(x-a_1)(x-a_3)+(x-a_2)(x-a_4)\\ & f(-\mathrm{\infty })=\mathrm{\infty }\\ & f\left(a_1\right)=+ve\\ & f\left(a_2\right)=-ve\\ & f\left(a_3\right)=-ve\\ & f\left(a_4\right)=+ve\\ & f(\mathrm{\infty })=\mathrm{\infty }\end{array}$

$f{\left(a_1\right)}_{\mathrm{\&}}f\left(a_2\right)$ are of opposite signs, So at least one root between them. \& $f\left(a_3\right),f\left(a_4\right)$ are of opposite signs so at least one root between them. But being a quadratic can't have more than two roots, So exactly one in $(a_1,a_2)$ and exactly one in $(a_3,a_4)$ so both roots are real.

$\therefore$ Option (C)