I'm really not sure regarding this question. would like your help if possible.

given the function:

$ f(n) = \begin{cases} \frac{e^x-1}{x}, & \text{if $x \neq0$ } \\ 1 & \text{if $x=0$ } \end{cases} $

a) Prove for every $x\neq0$ $e^x(x-1)+1>0.$

b) Prove that $f$ in continuous in R.

c) Use the above to prove that $f$ is monotonously increasing in R.

d) Prove that every $c>0$ has a solution for $f(x)=c.$

What i did:

a) I used it as a function and tried to show that the limit is always positive, when $x \to\infty$ and $x\to\infty$ while showing that it doesn't hold for $x=0.$

b) Using the limits of a) to show continuity, with the exception of $x=0,$ which I didn't know how to prove continuous in regards to the other part of the function.

c) Didn't know how to prove it.

d)using the above, I tried to use the intermediate value theorem, by showing that since the function is continuous, then there most exist a $c$ so that $f(x)=c.$ I didn't know how to prove that only for $c>0.$

Please show me the right way to do it, as I'm sure that I've done too many errors, and I don't know how to do it correctly.

Thank you very much.


i will start with b to be continuous in R we see that (e^x+1)/x continuous in R{0} but f continuous in 0 because f(0)=1 lim (e^x+1)/x when x---->0 =1 then c we have to Find derivative f`(x)= (e^x)/(x^2) and =0 if x=0 f'>(or=0) then f is monotonously increasing d- Since the function is always increasing Each horizontal line will be offset by y = c At only one point is the solution i am very happy becaus i help you and if you found questoin ask me and i will tey in (a)

  • 1
    $\begingroup$ thank you very much for your help. how would you approach to solving the inequality $e^x(x-1)+1>0$ for every $x \neq 0$? thank again $\endgroup$ – BeginningMath Jul 6 '17 at 20:09
  • 1
    $\begingroup$ ok thank you i will try with a e^x+1/(x-1)>0 if (x-1)>0 and e^x+1/(x-1)<0 if (x-1)<0 and x isnt 0 then let start with e^x+1/(x-1)>0 if (x)>1 we can see e^x>1 and 1/(x-1)>1 then e^x+1/(x-1)>1>0 let we see in the second e^x+1/(x-1)<0 if (x)<1 we can see e^x<1 and 1/(x-1)<1 then e^x+1/(x-1)<0 thene^x+1/(x-1)>0 i hope i can help you $\endgroup$ – small Jul 6 '17 at 21:24
  • $\begingroup$ math.meta.stackexchange.com/questions/5020/…. $\endgroup$ – user21820 Aug 31 '17 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.