# How do I use the intermediate value theorem to determine whether x^5 + 3x^2 - 1 = 0 has a solution over the interval [0, 3]?

May 3, 2018

You need only use Bolzano's Theorem. See below

#### Explanation:

Bolzano's Theorem state that "If f continous over [a,b] and f(a)·f(b)<0, then there is c in (a,b) such that f(c)=0"

Note that f(a)f(b)<0 means that f(a) and f(b) have diferent sign in a and b

In our case $f \left(x\right) = {x}^{5} + 3 {x}^{2} - 1$ is a continous function (because is polinomical) and

$f \left(0\right) = - 1 < 0$
$f \left(3\right) = 269 > 0$

By Bolzano's theorem there is a value c between 0 and 3 such that f(c)=0

In this case (and in general) you can do a fine tunnig of c. For example
f(0)<0
f(1)=3>0

Then that c is between 0 and 1

We can`t assume that there is no other c between 1 and 3 because f has no sign change there...

May 3, 2018

#### Explanation:

The intermediate value theorem states :

If the function $f \left(x\right)$ is continuous on the interval $\left[a , b\right]$ and $y$ is a number between $f \left(a\right)$ and $f \left(b\right)$, then there exists a number $x = c$ in the open interval $\left(a , b\right)$ such that $y = f \left(c\right)$.

Here,

$f \left(x\right) = {x}^{5} + 3 {x}^{2} - 1$

is a continouos function on the interval $\left[0 , 3\right]$ as $f \left(x\right)$ is a polynomial function.

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

$f \left(3\right) = 243 + 27 - 1 = 269$

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

As $f \left(x\right)$ changes sign from $f \left(0\right)$ to $f \left(3\right)$

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

This is the application of the intermediate value theorem.