4)a) prove that gets a unique value for every x in R
4)b)prove that gets total minimum (not only local)
you can use part 4a)
4)c)prove that 'uniformly' continuous in
regarding a:
ok here is the derivative what should i do whith it.
i know that both the the function and the derivative are continues because they are combination of continues functions.
maybe i should take limits to infinity and dissasamble the result into definition
?
looking for guidance
what t do next
for what purpose
Show that the function is either 1-to-1 or many-to-1. Again, if this question is from a past test, exam etc. it is not our job to provide solutions to tests, exams etc. from a previous semester. It is the job of your institute to provide the solutions. Ask your instructors. If they are unwilling to provide the solutions, there is probably a very good reason for that .....
At least do the algebra to simplify that!
Now, look at the sign of that- where the function is increasing or decreasing.
ok here is the derivative what should i do whith it.
i know that both the the function and the derivative are continues because they are combination of continues functions.
maybe i should take limits to infinity and dissasamble the result into definition
?
looking for guidance
what t do next
for what purpose
its monotonicly increasing because the derivative is positive.
and there is no extreme points because i cant isolate x in f'(x)=0
so it could not be solved so its always increasing and for each x each have a uniques value.
correct?
regrading b):
i was told to use part a
so i did like in a:
i cant solve
there is no way to isolate x
how to prove that there is a total minimum if we cant find extreme points
??
another way is that i can say that its continues on some close subsection
so minimum must present there by weirshtrass
yes there are both continues because they are combination of continues functions
,but if you say that we dont have to look for it only to prove existance
then,the function is continues on every closed section so by weirshtrass we have a local minimum point,
but i was asked about total minimal point,so i dont know if its sufficient.
at f(0)=1 f(1)=e+0.5
so there t between the values for which f(t) gt value between 1 and e+0.5
Did you not notice that f'(x)= g(x)? And you have already proved that g is an increasing function. There can be only one value of x at which f'(x)= 0. And, you have proven that the derivative of g, so the second derivative of f, is positive for all x so particularly where f'(x)= g(x)= 0. What does knowing that f''(x) is positive tell you about the critical point?
that its minimum ,and because we we proved that g(x) has unique value for each x
that we f' got only one extreme point.thus making it global
thanks
but regarding the prove of a presence of extreme point.
why couldnt we use here the theorem of weirshtass where if f(x) is continues on closed section then there is minimal point inside.
?
(we have continuety on all R )