# Inverse Functions

• Aug 26th 2008, 09:45 AM
Coco87
Inverse Functions
Hey, I've just started calc 2, I have two problems that are driving me crazy, and need some help (Crying)

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):

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

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

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:
$\displaystyle f(x)=x^3+3\sin{x}+2\cos{x}$
$\displaystyle g(x)=f^{\neg1}(x)$
$\displaystyle g^\prime(2)=?$
$\displaystyle f^\prime(x)=3x^2+3\cos{x}-2\sin{x}$
$\displaystyle 2=x^3+3\sin{x}+2\cos{x}$

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

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

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

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

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

Thanks!
• Aug 26th 2008, 10:33 AM
Moo
Hi !
Quote:

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

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):

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

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

If you want to apply conjugates... :)

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

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

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

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

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

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

Quote:

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:
$\displaystyle f(x)=x^3+3\sin{x}+2\cos{x}$
$\displaystyle g(x)=f^{\neg1}(x)$
$\displaystyle g^\prime(2)=?$
$\displaystyle f^\prime(x)=3x^2+3\cos{x}-2\sin{x}$
$\displaystyle 2=x^3+3\sin{x}+2\cos{x}$

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:

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

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

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

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

Thanks!
Ok

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

Let's apply chain rule :

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

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

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

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 $\displaystyle x=0$, it works :
$\displaystyle 2=0^3+3\sin(0)+2\cos(0)=2$
Hence $\displaystyle g(2)=0$.

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

(Whew)
• Aug 26th 2008, 10:44 AM
o_O
If you need to be picky, note that the range of your original function is $\displaystyle -1 < y \leq 1$ so the domain of your inverse function will be $\displaystyle -1 < x \leq 1$.
• Aug 26th 2008, 11:02 AM
Coco87
Quote:

Originally Posted by Moo
Hi !

If you want to apply conjugates... :)

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

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

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

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

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

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

Haha, I guess I was way off on that one (Itwasntme) Seems to be my problem, not good at finding answers like that.

Wow, did I completely over analyze that (Surprised) I understand now.

Thanks!