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#

Answer 1

Christian Antunez

Check below.

I now understand.

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

We have two options:

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

At least one of the two numbers should be the opposite of the other.

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#

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Answer 2

Lincoln Anderson

See below.

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 accurate, if not

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

will suggest 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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.