Question

prove \(\tan ^{-1}x>\frac{x}{1+\frac{1}{3}x^2}\) where x>0

Answer 1

@mathstudent55

Answer 2

Any idea? I don't even know how to start solving this..

Answer 3

haven't come up with much myself :(

Answer 4

start with defining\[f(x)=\tan^{-1}x-\frac{x}{1+\frac{1}{3}x^2}\]show that \(f'(x)\) is always positive and note that \(f(0)=0\)

Answer 5

yeah, i get \(f^{1}x>0\) but how does that prove this thing?

Answer 6

What does \(f^1(x)\) mean?

Answer 7

First derivative of f(x)

Answer 8

Oh you meant \(f'(x)\) Haha,okay.

Answer 9

lol.. yeah.. i didn't know how to type that actually.. lol

Answer 10

Because when you find the first derivative to be positive, and f(0) = 0, you're essentially finding the max value.

Answer 11

Along with this. Hope it helps!

Answer 12

can i say that since \(f'(x)>0\) and f(x)=0, the nature of graph will be something like:

Answer 13

which implies that f(x)>0

Answer 14

I believe so.

Answer 15

thanks!

Answer 16

No problem :D