Question

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]`?

Answer 1

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(x)=x^5+3x^2-1` is a continous function (because is polinomical) and

`f(0)=-1<0`
`f(3)=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...

Answer 2

Please see the explanation below.

Explanation:

The intermediate value theorem states :

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

Here,

`f(x)=x^5+3x^2-1`

is a continouos function on the interval `[0,3]` as `f(x)` is a polynomial function.

`f(0)=-1`

`f(3)=243+27-1=269`

`0 in (f(0), f(3))`

As `f(x)` changes sign from `f(0)` to `f(3)`

`EE c in (0,3)` such that `f(c)=0`

This is the application of the intermediate value theorem.