# Thread: I need help with solving inequalities via derivatives

1. ## 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)

2. Originally Posted by ppvanya
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

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3. ## 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!

4. Originally Posted by ppvanya
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

5. Originally Posted by ppvanya
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)

$e^x\geqslant \ln(1+x)\implies e^{e^x}\geqslant 1+x$. 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.