unique value and extreme point question

4)a) prove that $\displaystyle g(x)=e^{x}+\frac{x^{3}}{x^{2}+1} $gets a unique value for every x in R

4)b)prove that $\displaystyle f(x)=e^{x}+\frac{x^{2}}{2}-\frac{ln(x^{2}+1)}{2}$ gets total minimum (not only local)

you can use part 4a)

4)c)prove that $\displaystyle \frac{ln(x+1)}{x}$ 'uniformly' continuous in $\displaystyle (0,\infty)$

Re: unique value and extreme point question

Well, what have **you** done on this? For (a), I recommend looking at the derivative. For (b), since it say "you can use part 4a", what is the **relationship** between the functions in (b) and (a)?

Re: unique value and extreme point question

regarding a:

$\displaystyle g'(x)=e^x+\frac{3x^2(x^2+1)-x^32x}{(x^2+1)^2}$

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

Re: unique value and extreme point question

Quote:

Originally Posted by

**transgalactic** regarding a:

$\displaystyle g'(x)=e^x+\frac{3x^2(x^2+1)-x^32x}{(x^2+1)^2}$

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 .....

Re: unique value and extreme point question

did you read this thread?

i was asked to prove that it gets unique value for every x in R ( 1 to 1),and i have presented a possible way of solution

could you see it

?

Re: unique value and extreme point question

Quote:

Originally Posted by

**transgalactic** regarding a:

$\displaystyle g'(x)=e^x+\frac{3x^2(x^2+1)-x^32x}{(x^2+1)^2}$

At least do the algebra to simplify that!

$\displaystyle g'(x)= e^x+ \frac{3x^4+ 3x^2- 2x^4}{(x^2+ 1)^2}= e^x+ \frac{x^4+ 3x^2}{(x^2+ 1)^2}$

Now, look at the sign of that- where the function is increasing or decreasing.

Quote:

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

Re: unique value and extreme point question

Quote:

Originally Posted by

**HallsofIvy** At least do the algebra to simplify that!

$\displaystyle g'(x)= e^x+ \frac{3x^4+ 3x^2- 2x^4}{(x^2+ 1)^2}$$\displaystyle = e^x+ \frac{x^4+ 3x^2}{(x^2+ 1)^2}$

Now, look at the sign of that- where the function is increasing or decreasing.

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:

$\displaystyle f'(x)=e^x +x-\frac{x}{x^2 +1}$

$\displaystyle f'(x)=e^x +\frac{x(x^2+1)-x}{x^2 +1}$

i cant solve $\displaystyle e^x +\frac{x(x^2+1)-x}{x^2 +1}=0$

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

Re: unique value and extreme point question

Simplify the numerator of the rational term.

Re: unique value and extreme point question

ok the derivative is $\displaystyle e^x +\frac{(x^3)}{x^2 +1}=0$

but i cant solve it,because its not possible to isolate x

so there is no extreme points

after simplifying like you said i cant see how the situation changed.

could you guide me about it

?

Re: unique value and extreme point question

Quote:

Originally Posted by

**transgalactic** ok the derivative is $\displaystyle e^x +\frac{(x^3)}{x^2 +1}=0$

but i can't solve it, because its not possible to isolate x

so there is no extreme points

after simplifying like you said i cant see how the situation changed.

could you guide me about it ?

Do you have to find the zero of a function to know that it has a zero?

Evaluate the first derivative at x = -1 and x = 0 & use the mean value theorem. (Are the function and its first derivative continuous?)

Re: unique value and extreme point question

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

Re: unique value and extreme point question

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?

Re: unique value and extreme point question

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 )