How do you verify the intermediate value theorem over the interval $[0,3]$, and find the $c$ that is guaranteed by the theorem such that $f\left( c \right) = 4$ where $f\left( x \right) = {x^3} - {x^2} + x - 2$?

Hint:To verify the intermediate value theorem, we will first find the value attained by the function at the starting and the ending of the given interval. If $4$ lies between those values, then we can say that there must exist a $c$ for which $f\left( c \right) = 4$. Then to find the value of $c$, we will apply Rational Zeros Theorem to find a list of all possible zeros of the equation where we equate the given function with $4$ and then we will use hit and trial method to find the value of $c$.

Complete step by step solution:
(i) We are given a function of $x$ i.e.,
$f\left( x \right) = {x^3} - {x^2} + x - 2$
As we know that the intermediate value theorem states that if a continuous function is capable of attaining two values for an equation, then it must also attain all the values that are lying in between these two values in the same interval.
In simpler words, since we know that $f\left( x \right)$ is a polynomial so it is continuous in the interval $[0,3]$.
So, according to the intermediate value theorem we can say for the function $f\left( x \right)$ in the interval $[0,3]$ that if the value of $f\left( 0 \right) < 4$ and the value of $f\left( 3 \right) > 4$ or the value of $f\left( 0 \right) > 4$ and the value of $f\left( 3 \right) < 4$, then there exists a point $c$ which has a value which is in the interval $[0,3]$ such that $f\left( c \right) = 4$
In order to verify it, we will find the value attained by the function at the starting point of the interval and at the ending point i.e., $f\left( 0 \right)$ and $f\left( 3 \right)$. Therefore,
$
f\left( 0 \right) = {0^3} - {0^2} + 0 - 2 \\
f\left( 0 \right) = - 2 \\
$
And,
$
f\left( 3 \right) = {3^3} - {3^2} + 3 - 2 \\

f\left( 3 \right) = 27 - 9 + 3 - 2 \\
f\left( 3 \right) = 19 \\
$
Now, because the value of $f\left( 0 \right) < 4$ and the value of $f\left( 3 \right) > 4$ and the function $f\left( x \right)$ is continuous in the closed interval $[0,3]$, the expression satisfies the conditions of the intermediate value theorem.
 Therefore, there exists a point $c$ in the interval $[0,3]$ for which $f\left( c \right) = 4$.
(ii)In order to find the value of $c$, we will apply the Rational Zeros Theorem to the function $f\left( x \right) = {x^3} - {x^2} + x - 2$ when equated with $4$ and find the list of possible zeros of the equation.

Therefore, we will first equate the function with $4$.
${x^3} - {x^2} + x - 2 = 4$

Subtracting $4$ from both the sides, we will get:
$
{x^3} - {x^2} + x - 2 - 4 = 4 - 4 \\
{x^3} - {x^2} + x - 6 = 0 \\
$
Now we will apply Rational Zeros Theorem to the equation.

Since, the constant is $ - 6$, its factors would be $ \pm 1, \pm 2, \pm 3, \pm 6$ and since the leading coefficient is $1$, its factors would be $
\pm 1$.

Dividing the factors of $ - 6$ by the factors of $1$, we will get the following list of possible zeros:
$ \pm 1, \pm 2, \pm 3, \pm 6$

(iii) As we have the list of all possible zeros, we will put all these values in the function and test one by one.
After testing, it shows that $2$ is a solution as
${2^3} - {2^2} + 2 - 6 = 8 - 4 + 2 - 6 = 0$
Therefore, $c = 2$
And,
$
f\left( c \right) = {2^3} - {2^2} + 2 - 2 \\
f\left( c \right) = 8 - 4 + 2 - 2 \\
f\left( c \right) = 4 \\
$

Hence, for $f\left( x \right) = {x^3} - {x^2} + x - 2$ there exist $c$ in the interval $[0,3]$ such that $f\left( c \right) = 4$ i.e., $c = 2$.
Note: A function is termed continuous when its graph is an unbroken curve. Since, $f\left( x \right)$ is a polynomial cubic function, it is continuous at each point in the given interval.
Also, the second part of the question where we found the value of $c$, we could also have just applied hit and trial method instead of Rational Zeros Theorem by taking $x$ as $ \pm 1, \pm 2$ because these are the most common roots we obtain in such type of equations.