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]#?

2 Answers
May 3, 2018

Answer:

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...

May 3, 2018

Answer:

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.