# Math Help - Inverse Functions

1. ## Inverse Functions

Hey, I've just started calc 2, I have two problems that are driving me crazy, and need some help

The first one is probably extremely simple, but I've tried pretty much everything I can think of. The problem is that I need to find the inverse function of f(x):

$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}$

I remember a similar problem involving conjugates, but I can't figure out how to apply it correctly (or I'm retarded ).

The second one is an example our instructor gave us, but I can't figure out where she got the "0" for cos and sin -

With this problem, we try to find the derivative of the inverse of f(x) where g(x) is the inverse:
$f(x)=x^3+3\sin{x}+2\cos{x}$
$g(x)=f^{\neg1}(x)$
$g^\prime(2)=?$
$f^\prime(x)=3x^2+3\cos{x}-2\sin{x}$
$2=x^3+3\sin{x}+2\cos{x}$

Now here's the part where she skipped a lot of stuff and literally put this on the board:

$\cos{0}=1$
$\sin{0}=0$

$g^\prime(2)=\frac{1}{3}$

Where did she get "0" from? I have no idea where that fits into the problem.

Any help I can get would be awesome, even if it's just a hint.

Thanks!

2. Hi !
Originally Posted by Coco87
Hey, I've just started calc 2, I have two problems that are driving me crazy, and need some help

The first one is probably extremely simple, but I've tried pretty much everything I can think of. The problem is that I need to find the inverse function of f(x):

$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}$

I remember a similar problem involving conjugates, but I can't figure out how to apply it correctly (or I'm retarded ).
If you want to apply conjugates...

The conjugate of $a+b \sqrt{c}$ is $a-b \sqrt{c}$. The product of the two is $(a+b \sqrt{c})(a-b \sqrt{c})=a^2-b^2c$ by the identity $(x-y)(x+y)=x^2-y^2$.

$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}$ multiply by $\frac{1-\sqrt{x}}{1-\sqrt{x}}=1$ :

$f(x)=y=\frac{(1-\sqrt{x})^2}{1-x}$

But I think it won't help ^^'
---------------------------
$y=\frac{1-\sqrt{x}}{1+\sqrt{x}} \implies y+y \sqrt{x}=1-\sqrt{x}$

$\sqrt{x}(y+1)=1-y$

$\sqrt{x}=\frac{1-y}{1+y} \implies x=\left(\frac{1-y}{1+y}\right)^2=\boxed{f^{-1}(y)}$

The second one is an example our instructor gave us, but I can't figure out where she got the "0" for cos and sin -

With this problem, we try to find the derivative of the inverse of f(x) where g(x) is the inverse:
$f(x)=x^3+3\sin{x}+2\cos{x}$
$g(x)=f^{\neg1}(x)$
$g^\prime(2)=?$
$f^\prime(x)=3x^2+3\cos{x}-2\sin{x}$
$2=x^3+3\sin{x}+2\cos{x}$

Now here's the part where she skipped a lot of stuff (SiMoon says : I don't think she did, you'll see below ) and literally put this on the board:

$\cos{0}=1$
$\sin{0}=0$

$g^\prime(2)=\frac{1}{3}$

Where did she get "0" from? I have no idea where that fits into the problem.

Any help I can get would be awesome, even if it's just a hint.

Thanks!
Ok

$x=f(f^{-1}(x))=f(g(x))$

Let's apply chain rule :

$1=g'(x)\cdot f'(g(x))$

Therefore $g'(2)=\frac{1}{f'(g(2))}$

We have to calculate g(2), that is to say x such that $2=x^3+3\sin{x}+2\cos{x}$, because $g(2)=f^{-1}(2) \implies 2=f(x)$

What your teacher did was to note that there is a 2 in the LHS. There's a coefficient 2 in the RHS. And magically, if you put $x=0$, it works :
$2=0^3+3\sin(0)+2\cos(0)=2$
Hence $g(2)=0$.

Therefore $g'(2)=\frac{1}{f'(g(2))}=\frac{1}{f'(0)}=\frac 13$

3. If you need to be picky, note that the range of your original function is $-1 < y \leq 1$ so the domain of your inverse function will be $-1 < x \leq 1$.

4. Originally Posted by Moo
Hi !

If you want to apply conjugates...

The conjugate of $a+b \sqrt{c}$ is $a-b \sqrt{c}$. The product of the two is $(a+b \sqrt{c})(a-b \sqrt{c})=a^2-b^2c$ by the identity $(x-y)(x+y)=x^2-y^2$.

$f(x)=\frac{1-\sqrt{x}}{1+\sqrt{x}}$ multiply by $\frac{1-\sqrt{x}}{1-\sqrt{x}}=1$ :

$f(x)=y=\frac{(1-\sqrt{x})^2}{1-x}$

But I think it won't help ^^'
---------------------------
$y=\frac{1-\sqrt{x}}{1+\sqrt{x}} \implies y+y \sqrt{x}=1-\sqrt{x}$

$\sqrt{x}(y+1)=1-y$

$\sqrt{x}=\frac{1-y}{1+y} \implies x=\left(\frac{1-y}{1+y}\right)^2=\boxed{f^{-1}(y)}$
Haha, I guess I was way off on that one Seems to be my problem, not good at finding answers like that.

Originally Posted by Moo
Ok

Let's apply chain rule :

Therefore

We have to calculate g(2), that is to say x such that , because

What your teacher did was to note that there is a 2 in the LHS. There's a coefficient 2 in the RHS. And magically, if you put , it works :

Hence .

Therefore

Wow, did I completely over analyze that I understand now.

Thanks!