Prove using mean-value theorem that $x/(1+x^2)<\arctan x
I got the first part but how do I prove $\arctan x< x$ using the MVT?
The first part was done easily by applying MVT on $\arctan x$, should I use $\arctan x-x$ for the second part? Thanks!
Prove using mean-value theorem that $x/(1+x^2)<\arctan x
I got the first part but how do I prove $\arctan x< x$ using the MVT?
The first part was done easily by applying MVT on $\arctan x$, should I use $\arctan x-x$ for the second part? Thanks!
I think you have it. One application of the MVT gives you both parts:
Let $x>0$. Applying the Mean Value Theorem to $f(x)=\arctan x$ on the interval $[0,x]$ gives a number $c$ with $0
Since $x>c$ and $x\gt0$, we have: ${x\over 1+x^2}\lt{x\over 1+c^2}