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Math Help - I need help with solving inequalities via derivatives

  1. #1
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    I need help with solving inequalities via derivatives

    Hello guys, I have a problem with solving these inequalities. If anyone could help me I would really appreciate it.

    1. e^x>1+ln(1+x), for all real numbers ( I understand how this is true for x>0, but somehow I can't seem to make it work for x<0, although it is obviously true, since f'(x)<0 for all x's except x=0.

    2. sinx<x- (x^3)/6+(x^5)/120, for x>=0 (this one doesn't seem quite right, since its graph kind of stays constant around x=2.5-3. Main question is how do I tell whether the derivative is positive or negative when it is a complicated function itself and it's hard to plug in numbers and check)

    Thank you in advance!
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  2. #2
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    Quote Originally Posted by ppvanya View Post
    Hello guys, I have a problem with solving these inequalities. If anyone could help me I would really appreciate it.

    1. e^x>1+ln(1+x), for all real numbers ( I understand how this is true for x>0, but somehow I can't seem to make it work for x<0, although it is obviously true, since f'(x)<0 for all x's except x=0.


    We define f(x)=e^x-\ln(x+1)-1 . Since \lim_{x\to -1}f(x)=\infty\,,\,\, f(0)=0\,\,\,and\,\,\,f'(x)=e^x-\frac{1}{x+1}=0\Longleftrightarrow x=0 (why? Because f'(x) is an increasing function) , we have that our function has a minimal point at x =0 and thus \forall\,x\in (-1,\infty)\,,\,\,f(x) = e^x-\ln(x+1)-1\ge 0 = f(0)

    2. sinx<x- (x^3)/6+(x^5)/120, for x>=0 (this one doesn't seem quite right, since its graph kind of stays constant around x=2.5-3. Main question is how do I tell whether the derivative is positive or negative when it is a complicated function itself and it's hard to plug in numbers and check)

    For this one you better check the power series (Taylor-MacClaurin polynomial) around zero of \sin x.

    Tonio

    Thank you in advance!
    .
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  3. #3
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    Must use derivatives, not limits :(

    Thanks for the reply, but this must be solved by using derivatives only. Here's what I did (And I'm really sorry I don't know how to write it in the fancy way you do)

    e^x>1+ln(1+x), for all real numbers
    I did this f(x)=e^x-1-ln(1+x), then f'(x)=e^x-1/(1+x). Since I have to prove the inequality for all real numbers, I look at x<0, so I plug in some negative number for x in f'(x). For example x=-7, and f'(-7)>0, which means that for x<0, f(x) is increasing. But check the graph! It's decreasing for the negative x's!!!!! I am stumped.... Why is the derivative +, when the function is actually decreasing?

    Thanks again!
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  4. #4
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    Quote Originally Posted by ppvanya View Post
    Thanks for the reply, but this must be solved by using derivatives only. Here's what I did (And I'm really sorry I don't know how to write it in the fancy way you do)

    e^x>1+ln(1+x), for all real numbers
    I did this f(x)=e^x-1-ln(1+x), then f'(x)=e^x-1/(1+x). Since I have to prove the inequality for all real numbers, I look at x<0, so I plug in some negative number for x in f'(x). For example x=-7, and f'(-7)>0, which means that for x<0, f(x) is increasing. But check the graph! It's decreasing for the negative x's!!!!! I am stumped.... Why is the derivative +, when the function is actually decreasing?

    Thanks again!

    First, my solution and the Taylor polynomial proposal use derivatives only, second your function is defined ONLY in (-1,\infty) , so evaluating its derivative in -7 is pretty nonsensical.
    Third, the function's always increasing.

    Tonio
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  5. #5
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    Quote Originally Posted by ppvanya View Post
    Hello guys, I have a problem with solving these inequalities. If anyone could help me I would really appreciate it.

    1. e^x>1+ln(1+x), for all real numbers ( I understand how this is true for x>0, but somehow I can't seem to make it work for x<0, although it is obviously true, since f'(x)<0 for all x's except x=0.

    2. sinx<x- (x^3)/6+(x^5)/120, for x>=0 (this one doesn't seem quite right, since its graph kind of stays constant around x=2.5-3. Main question is how do I tell whether the derivative is positive or negative when it is a complicated function itself and it's hard to plug in numbers and check)

    Thank you in advance!
    e^x\geqslant \ln(1+x)\implies e^{e^x}\geqslant 1+x. But, clearly for all real numbers e^{e^x}\geqslant e^x so it remains to show that e^x\geqslant 1+x. This can be done fairly easily.
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