Question

No graphing. Do functions intersect?
(some ideas)

Answer 1

Hello!
For now, lets suppose there are some two functions \(\rm F\) and \(\rm G\),
that are continuous over (-∞,+∞). The question I will address is
whether these functions are going to intersect.

This will involve \(\rm calc~I.\) I don't plan to involve more advanced math....
it should be easy to read.

Answer 2

Lets now also suppose suppose that one of the functions increases and
the other decreases on the interval \(\rm N\). For example, \(\rm F\) increases, and \(\rm G\)
decreases (over the interval \(\rm N\).

(since these functions are continuous over (-∞,+∞), they're continuous on \(\rm N\).)

Answer 3

If ranges (over N) of the functions intersect (keeping in mind the previous conditions I proposed), then the functions also intersect, and if the ranges (over N) don't intersect then the functions for sure don't intersect (over N).

Again, the range (over N) of these functions is from it's absolute maximum (over N) to absolute minimum (over N). And keep in mind they are cont. over (-∞,+∞)

the way to find maximum and minimum is to find critical numbers
and check for f(what)=smallest and f(what)=largest
((the range is from smallest to largest output.))

so lets say that for F the max=B min=A
and for G the max=D and min=C

so range of F = [A,B]
and range of G=[C,D]

then if ranges intersect, F and G intersect.

if ranges don't intersect (such that A>D, or such that C>B)
then F and G don't intersect.

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Note: for functions NOT to intersect, F and G won't intersect if ranges don't
intersect, even in a case when F and G both increase or both decrease. The
only times that F should increase and G decrease or vice versa, is to show that
they do intersect.

Answer 4

When I have learned calc I, I haven't learned any such proves, but when I saw this question I right away thought of this idea.

It is originated from \(\large {\bbox[5pt, lightcyan ,border:2px solid black ]{ \bf\color{blue }{\href{http:///openstudy.com/study#/updates/5582dd23e4b091b59af0b36e}{here}}}}\) .