Question

I'm trying to wrap my head around to prove that arctan is an odd function. In the assignment it is said that:

f(0) = 0 and

f'(x) = 1/(1+x^2)


I am then said that:
g(x) = f(-x) + f(x)

and I am supposed to derive g.

I find:

g'(x) = 2/(1+x^2)

which kind of seems wrong to me as I am supposed to find 2/(2+x^2)

Anyone got an idea?

Answer 1

integrate f' from -x to x

Answer 2

Well I would love to, but we are not supposed to do that. ( and I don't know how either yet )

The teacher left a note saying that z(x) = f(ax + b ) means that z'(x) = a * f'(ax+b)

hence why I want to find 2/(2+x^2) because then I can find 2 * f'(2x+b)
...
which actually doesn't make sense because I am supposed to find that their sum is 0 and not two. I am kind of at a loss here...

Answer 3

youll have to prolly ask your teacher that then, cause I got no idea what they are asking you to "do" then.

Answer 4

What is written there exactly is:

f(x) = arctan(x)

f(0) = 0
f'(x) = 1/(1+x^2)

z(x) = f(-x) + f(x)

and the question I am trying to answer is: "Find z'(x) in R"

Answer 5

if by z'(x) you mean taking the derivative of z(x); then thats just the chain rule

Answer 6

z'(x) = -f'(-x) + f'(x)

Answer 7

replace f' with its definition to get:
\[z'(x) = -\frac{1}{1+(-x)^2}+\frac{1}{1+x^2}=0\]

Answer 8

where did the -f'(x) come from though? I am not sure why it gets a negative sign all of the sudden.

Answer 9

the chain rule of derivatives

Answer 10

\[f(g(x))=f'(g(x))*g'(x)\]
g(x) = -x inthe first term; and x in the second term

Answer 11

the derivative of -x is -1; the derivative of x is 1

Answer 12

Oooh... that clears up things a bit.

If you don't mind helping me just a bit more, how did you put:

f(g(x))=f′(g(x))∗g′(x)
in there?

Is it because g(x) = x? Thus making it f'(-x) = f'(-x)*-1?

Answer 13

that last sentence is correct; and are you asking about using latex codes?

Answer 14

No, I didn't mean LaTeX I meant how you used the chain rule. But since you confirmed that my last sentence was correct it answered my question!
Thanks alot!

Answer 15

ok, your welocme :)

Answer 16

im still at a loss how that help to establish that inverse tangent is odd :/

Answer 17

if you can prove that z'(x) = 0 that means that z is constant.
as z(x) = f(-x) + f(x) and we know that f <=> arctan

we can show that z(x) = 0 thus proving that arctan is odd. ( This is how I see it, I hope I make sense )

Answer 18

if you can prove that z'(x) = 0; for all values of x; that means that z is constant.

Answer 19

That is what I said right...?

Answer 20

close, you forgot to include the "for all xs" part. z'(x)=0 implies that we have at least a min or max for some value of x

Answer 21

Ooh. Right! Yep, I shouldn't forget that. Still getting back into maths after summer break :P

Well thanks for the correction! When writing my report about it I will make sure to not forget it.

Answer 22

good luck ;)