Show that #f# has at least one root in #RR# ?

Given #f:RR->RR#, continuous with #f(a)+f(b)+f(c)=0# , #a,b,c# #in##RR#

2 Answers
Dec 12, 2017

Check below.

Explanation:

Got it now.

For #f(a)+f(b)+f(c)=0#

We can either have

  • #f(a)=0# and #f(b)=0# and #f(c)=0# which means that #f# has at least one root, #a# ,#b# ,#c#

  • One of the two numbers at least to be opposite between them

Let's suppose #f(a)=##-f(b)#
That means #f(a)f(b)<0#

#f# continuous in #RR# and so #[a,b]subeRR#

According to Bolzano's theorem there is at least one #x_0##in##RR# so #f(x_0)=0#

Using Bolzano's theorem in other intervals #[b,c]# ,#[a,c]# will lead to the same conclusion.

Eventually #f# has at least one root in #RR#

Dec 12, 2017

See below.

Explanation:

If one of #f(a), f(b),f(c)# equals zero, there we have a root.

Now supposing #f(a) ne 0, f(b) ne 0, f(c) ne 0# then at least one of

#f(a)f(b) < 0#
#f(a)f(c) < 0#
#f(b)f(c) < 0#

will be true, otherwise

#f(a)f(b) > 0, f(a)f(c) > 0, f(b)f(c) > 0#

will imply that

#f(a) > 0, f(b) > 0, f(c) > 0# or #f(a) < 0, f(b) < 0, f(c) < 0#.

In each case the result for #f(a)+f(b)+f(c)# could not be null.

Now if one of #f(x_i)f(x_j) > 0# by continuity, exists a #zeta in (x_i,x_j)# such that #f(zeta) = 0#