# 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(c)=4 where f(x)=x^3-x^2+x-2?

Aug 14, 2017

$f$ is a polynomial so $f$ is continuous on the interval $\left[0 , 3\right]$.

$f \left(0\right) = - 2$ and $f \left(3\right) = 19$

$4$ is between $f \left(0\right)$ and $f \left(3\right)$ so IVT tells us that there is a $c$ in $\left(0 , 3\right)$ with $f \left(c\right) = 4$

Finding the $c$ requires solving

${x}^{3} - {x}^{2} + x - 2 = 4$. Which is equivalent to

${x}^{3} - {x}^{2} + x - 6 = 0$

Possible rational zeros are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.

Testing shows that $2$ is a solution.

So $c = 2$

Aug 14, 2017

The value of $c \in \left(0 , 3\right)$ and is $c = 2$

#### Explanation:

The intermediate value theorem states that if $f \left(x\right)$ is a continuous function on the interval $\left[a , b\right]$, then there is a number $p$ between $f \left(a\right)$ and $f \left(b\right)$, $\left(f \left(a\right) \ne f \left(b\right)\right)$ such that there is a number $c \in \left(a , b\right)$ such that $p = f \left(c\right)$

Here, $f \left(x\right) = {x}^{3} - {x}^{2} + x - 2$, which is a polynomial function continuous on the Interval $\left[0 , 3\right]$

$f \left(0\right) = 0 - 0 + 0 - 2 = - 2$

$f \left(3\right) = {3}^{3} - {3}^{2} + 3 - 2 = 27 - 9 + 3 - 2 = 19$

$f \left(c\right) \in \left(f \left(0 , f \left(3\right)\right)\right)$

$f \left(c\right) = 4$ such that $c \in \left(0 , 3\right)$

Therefore,

$f \left(c\right) = {c}^{3} - {c}^{2} + c - 2 = 4$

$\implies$, ${c}^{3} - {c}^{2} + c - 6 = 0$

$f \left(2\right) = {2}^{3} - {2}^{2} + 2 - 2 = 8 - 4 + 2 - 2 = 4$