# How do you verify the intermediate value theorem over the interval [0,5], and find the c that is guaranteed by the theorem such that f(c)=11 where f(x)=x^2+x-1?

Sep 4, 2017

The value of $c = 3$

#### Explanation:

The Intermediate Value Theorem states that if $f \left(x\right)$ is a continuous function on the interval $\left[a , b\right]$ and $N \in \left(f \left(a\right) , f \left(b\right)\right)$, then there exists $c \in \left[a , b\right]$ such that $f \left(c\right) = N$

Here,

$f \left(x\right) = {x}^{2} + x - 1$ is continuous on $\mathbb{R}$ as it is a polynomial function.

The interval is $I = \left[0 , 5\right]$

$f \left(0\right) = - 1$

$f \left(5\right) = 25 + 5 - 1 = 29$

$f \left(x\right) \in \left[- 1 , 29\right]$

Then,

$f \left(c\right) = 11$ , $\implies$ $11 \in \left[- 1 , 29\right]$

$\exists c \in \left[0 , 5\right]$ such that $f \left(c\right) = 11$

Therefore,

${c}^{2} + c - 1 = 11$, $\implies$, ${c}^{2} + c - 12 = 0$

$\left(c + 4\right) \left(c - 3\right) = 0$

$c = - 4$ and $c = 3$

$c = 3$ and $c \in \left[0 , 5\right]$