Question

How do you use the Intermediate Value Theorem to show that the polynomial function $f\left( x \right) = \dfrac{{{x^3}}}{2} - 4x + \dfrac{1}{x}$ has a root in the interval $\left[ { - 3, - 1} \right]$?

Answer 1

To solve this question, we need to know the intermediate value theorem first. This theorem is about continuous function. It states that if $f\left( x \right)$ is a continuous function over a closed interval $\left[ {a,b} \right]$ with its domain having values $f\left( a \right)$ and $f\left( b \right)$at the endpoints of the interval, then the function takes any value between the values $f\left( a \right)$ and $f\left( b \right)$ at a point inside the interval.

Complete step by step solution:
Our first step here is to find the values of the function at $x = - 3$ and $x = - 1$.
We are given the function $f\left( x \right) = \dfrac{{{x^3}}}{2} - 4x + \dfrac{1}{x}$.
Let us put $x = - 3$ in the given equation.
$\Rightarrow f\left( { - 3} \right) = \dfrac{{{{\left( { - 3} \right)}^3}}}{2} - 4\left( { - 3} \right) + \dfrac{1}{{\left( { - 3} \right)}} = - \dfrac{{27}}{2} + 12 - \dfrac{1}{3} = - \dfrac{{11}}{6} < 0$
Now we will put $x = - 1$in the given equation.
$\Rightarrow f\left( { - 1} \right) = \dfrac{{{{\left( { - 1} \right)}^3}}}{2} - 4\left( { - 1} \right) + \dfrac{1}{{\left( { - 1} \right)}} = - \dfrac{1}{2} + 4 - 1 = \dfrac{5}{2} > 0$
From these values we can say that at one point, the function has a negative value and at the other point, it has a positive value. Therefore, it is obvious that at the first point, the curve is below zero and at the second point it is above zero.
Since the given equation is a polynomial, its graph will be continuous.
Thus, applying the intermediate value theorem, we can say that the graph must cross at some point between the given interval.
Hence, there exists a root of the function $f\left( x \right) = \dfrac{{{x^3}}}{2} - 4x + \dfrac{1}{x}$ in the interval $\left[ { - 3, - 1} \right]$.

Note: The intermediate value theorem which we have used here, has many applications. Mathematically, it is used in many areas. For example, this theorem is utilized to prove that there exists a point below or above a given particular line. It is also used to analyze the continuity of a function that is continuous or not.