Functions

**nath_quam** · July 2nd 2006, 04:27 PM

A function
$ y = f(x) $
under what geometrical conditions would
$ f(x) = f^{-1} (x) ? $
and whats an example of this function for which
$ f(x) = f^{-1} (x) ? $

&

does
$ \int^{1}_{-1} \sqrt$4-x^{2}$ $

$ = 2{\pi} $

**ThePerfectHacker** · July 2nd 2006, 07:34 PM

Originally Posted by nath_quam

A function
$ y = f(x) $
under what geometrical conditions would
$ f(x) = f^{-1} (x) ? $

For example, the identity function,
$f(x)=x$

---
Quick I am proud that yiu deleted your post

(I am moderator I can see everything you do, in case you are wondering).

**ThePerfectHacker** · July 2nd 2006, 07:37 PM

Originally Posted by nath_quam

$ \int^{1}_{-1} \sqrt$4-x^{2}$ $

$ = 2{\pi} $

Yes, because,
$y=\sqrt{4-x^2}$ is a semi-circle.
From $-2\leq x\leq 2$ the area below is,
$\frac{1}{2}\pi (2)^2=2\pi$

**Soroban** · July 2nd 2006, 08:02 PM

Hello, Nath!

Given a function: $y = f(x)$
(a) under what geometrical conditions would $f(x) = f^{-1}(x)\;?$

(b) What's an example of this function for which: $f(x) = f^{-1} (x)\:?$

(a) The graph must be symmetric to the line $y = x$ .

(b) A straight line like: . $y \;= \; 3 - x$
. . .A hyperbola like: . $y \:= \:\frac{1}{x}$

**nath_quam** · July 2nd 2006, 08:15 PM

Would someone be able to solve this

$ \int^{1}_{-1}\sqrt$4-x^{2}$ $
using
$ x = 2sin\theta\ $

**Soroban** · July 3rd 2006, 09:02 AM

Hello, Nath!

$\int^1_{-1}\!\sqrt{4-x^2}\,dx$ . using $ x = 2\sin\theta$

We can do it, but it takes a lot of work
and you're expected to be very familiar with Trigonometry.

I'll go through the whole thing in baby-steps . . .

Let $\x = 2\sin\theta\quad\Rightarrow\quadd x = 2\cos\theta\,d\theta$

Note that: . $\sqrt{4 - x^2} \:= \:\sqrt{4 - 2\sin^2\theta} \:=$ $\sqrt{4(1 - \sin^2\theta)} \:=\:\sqrt{4\cos^2\theta} \:=\:2\cos\theta$

Substitute: . $\int(2\cos\theta)(2\cos\theta\,d\theta) \:=\:4\int\cos^2\theta\,d\theta$

Double-angle identity: . $4\int\frac{1 + \cos2\theta}{2}\,d\theta \;= \;2\int(1 + \cos2\theta)\,d\theta$

Integrate: . $2\left[\theta + \frac{1}{2}\sin2\theta\right]$

Half-angle identity: . $2\left[\theta + \frac{1}{2}(2\sin\theta\cos\theta)\right] \:= \:2\left[\theta + \sin\theta\cos\theta\right]$

Now we must back-substitute.
We have: . $x = 2\sin\theta\quad\Rightarrow\quad\sin\theta = \frac{x}{2}\quad\Rightarrow\quad\theta = \arcsin\frac{x}{2}$
. . and: . $\cos\theta = \frac{\sqrt{4-x^2}}{2}$

So we have: . $2\left[\,\arcsin\frac{x}{2} + \frac{x\sqrt{4-x^2}}{4}\,\right]^1_{\text{-}1}$

$2\left[\arcsin\left(\frac{1}{2}\right) + \frac{1\cdot\sqrt{4-1^2}}{4}\right] -$ $2\left[\arcsin\left(-\frac{1}{2}\right) + \frac{(-1)\sqrt{4 - (-1)^2}}{4}\right]$

. . $= \;2\left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right) - 2\left(-\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$

. . $= \;\frac{\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

. . $= \;\frac{2\pi}{3} + \sqrt{3} \;\approx\;3.826446$

**nath_quam** · July 3rd 2006, 05:33 PM

Originally Posted by Soroban

Integrate: . $2\left[\theta + \frac{1}{2}\sin2\theta\right]$

At this stage couldn't you x sub 1 and -1 into
$ x = 2\sin\theta\ $
and get the answer that way
so you end up with
$ 2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}} $

**Soroban** · July 3rd 2006, 09:10 PM

hello, Nath!

Your idea is a good one . . .

At this stage couldn't you x sub 1 and -1 into $x = 2\sin\theta$ and get the answer that way . . . Yes!

so you end up with: $2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}$ . . . but your limits are incorrect.

We have: . $2\sin\theta = x$

For $x = -1:\;\;2\sin\theta\,=\,-1\quad\Rightarrow\quad\sin\theta\,=\,-\frac{1}{2}\quad\Rightarrow\quad\theta = -\frac{\pi}{6}$

For $x = 1:\;\;2\sin\theta\,=\,1\quad\Rightarrow\quad\sin \theta\,=\,\frac{1}{2}\quad\Rightarrow\quad \theta = \frac{\pi}{6}$

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