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]#?

Question

Chloe Alexander · Accepted Answer

You need only use Bolzano's Theorem. See below

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...

Madelyn Anderson · Answer

Please see the explanation below.

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.