Question

How do you use the intermediate value theorem to verify that there is a zero in the interval [0, 1] for $f\left( x \right)={{x}^{3}}+3x-2$ ?

Answer 1

Hint: In this equation, we have to verify that there is a zero in the closed interval [0, 1]. Thus, we will use the intermediate theorem to prove the problem. As we know, the intermediate value theorem states that if f is a continuous function in the interval [a, b] and c is any number between f (a) and f (b), then there is a number x in the closed interval such that$f\left( x \right)=c$ . So, we start solving this problem, by substituting the value 0 and 1 in place of x, and thus, we get the value of $f\left( 0 \right)$ and $f\left( 1 \right)$ . Since, it is given to us that $c=0$ , therefore we will verify that 0 will lie in between $f\left( 0 \right)$ and $f\left( 1 \right)$ , which is the required solution for the problem.

Complete step by step solution:
According to the problem, we have to prove the given statement of the function in the problem.
Thus, we will use the intermediate value theorem to get the solution.
The function given to us is $f\left( x \right)={{x}^{3}}+3x-2$ -------- (1)
So, for the intermediate value theorem, we know that the function must be continuous at the closed intervals. Therefore, we know that the function given to us is a polynomial function, and all polynomial functions are continuous at closed intervals $\left[ a,b \right]$ . Thus, we get
$f\left( x \right)={{x}^{3}}+3x-2$ is a continuous function at $\left[ 0,1 \right]$
Now, we will use the substitution method and will put the value of a and b in the place of x, that is let $a=0=x$ , now we will put this value in equation (1), we get
$\Rightarrow f\left( 0 \right)={{0}^{3}}+3\left( 0 \right)-2$
On further simplification, we get
$\Rightarrow f\left( 0 \right)=-2$ ---------- (2)
Now, we will again substitute the value of $b=1=x$ in equation (1), we get
$\Rightarrow f\left( 1 \right)={{1}^{3}}+3\left( 1 \right)-2$
Therefore on further solving, we get
$\Rightarrow f\left( 1 \right)=1+3-2$
Thus, we get
$\Rightarrow f\left( 1 \right)=2$ --------- (3)
So, we know that the value of c is in the given problem is $c=0$ , thus we can say that c lies between f (a) and f (b), that is
$f\left( a \right)\le c\le f\left( b \right)$
$f\left( 0 \right)\le c\le f\left( 1 \right)$
Now, we will substitute the value of equation (2), (3) and value of c in the above expression, we get
$\Rightarrow -2\le 0\le 2$
The above expression is true, thus the intermediate value theorem is applicable.
Therefore, there exists a number x such that $f\left( x \right)=c$ .
Thus, we have proved that using the intermediate value theorem, their exist a number c which is equal to 0, such that zero lies in between the interval [0, 1] of the function $f\left( x \right)={{x}^{3}}+3x-2$

Note: While solving this problem, do mention all the steps properly to avoid confusion and errors. Always mention the intermediate value theorem and its statement to get an accurate answer. Do not forget to prove that the function is continuous, because if it is not then we cannot apply the intermediate value theorem.