Hello everyone Im new to the forum but looking foward to getting to know everyone. Im a first year calculus desperately needing help to solve this problem. I have to prove thie following inequality using the mean value theorem;

a-b <=sin (b)-sin (a)<=b-a

Originally Posted by sicbagger
Hello everyone Im new to the forum but looking foward to getting to know everyone. Im a first year calculus desperately needing help to solve this problem. I have to prove thie following inequality using the mean value theorem;

$\displaystyle a-b\le\sin (b)-\sin (a)\le b-a$
You know that $\displaystyle \sin(x)$ is differentiable on every $\displaystyle [a,b]$.

You know that $\displaystyle \forall x,~-1\le \cos(x)\le 1$

Apply the mean value theorem to $\displaystyle \sin(x)$ on $\displaystyle [a,b]$

Thank you for the reply but this is my first semester of calc. I feel great that I understand your first sentence but your second sentence definitely lost me. Lol

Originally Posted by sicbagger
Thank you for the reply but this is my first semester of calc. I feel great that I understand your first sentence but your second sentence definitely lost me. Lol
Are you saying that you do not understand $\displaystyle \forall x,~-1\le \cos(x)\le 1~?$

If you are, then you do not understand enough Pre-Calculus to do this question.

I dont understand what the symbol in front of your x that looks like an upside down A is. I understand what -1 <=cos (x)<=1 means

Originally Posted by sicbagger
I dont understand what the symbol in front of your x that looks like an upside down A is. I understand what -1 <=cos (x)<=1 means
An upside-down A,$\displaystyle ~\forall~$, reads "for all"; just as for a backwards E,$\displaystyle ~\exists\;\; ,$ reads "there exists".

The Mean Value Theorem: if $\displaystyle f$ is differentiable on $\displaystyle [a,b]$ then $\displaystyle \exists c\in (a,b)$ such that $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$.