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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.

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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)

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    $\begingroup$ 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
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    $\begingroup$ 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

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