relative maximum,Newton’s Method,Integrate.......

**m777** · December 21st 2006, 11:45 PM

please try to solve these questions.

**CaptainBlack** · December 22nd 2006, 01:50 AM

Originally Posted by m777

please try to solve these questions.

Question # 1
The derivative of the continuous function is given. Find all critical points and determines whether a relative maximum, relative minimum or neither occurs there

$f'(x)=2\,\sin^3(x) - \sin^2(x),\ \ \ 0<x<2\,\pi$

The critical points are solutions of:

$f'(x)=2\,\sin^3(x) - \sin^2(x)=0,\ \ \ 0<x<2\,\pi$

which occur either when $sin(x)=0$ or $2\,\sin(x)-1=0$ , the roots of these are:

$\pi$ for the first and $\pi/6$ and $5 \pi/6$ for the second.

Now these correspond to local maxima when $f''(x)<0$ , minima when $f''(x)>0$ and (stationary) points of inflection when $f''(x)=0$ .

$f''(x)=2\,\sin(x)\cos(x)[3\, \sin(x)-1]$

So the first of the roots gives $f''(x)=0$ , the second $f''(x)>0$ and the third $f''(x)<0$

RonL

**CaptainBlack** · December 22nd 2006, 02:07 AM

Originally Posted by m777

please try to solve these questions.

Question # 2
Solve it with the help of Newton’s Method up to four terms

$sin(x)=x^2$

We want to find the roots of: $f(x)=\sin(x)-x^2$ using Newton's method. This is the ittereation:

$x_{n+1}=x_n-f(x_n)/f'(x_n)$

and here: $f'(x)=\cos(x)-2x$ .

This can be set up in Excel (see attachment) from which we see that there is a root at x~=0.8767

**Soroban** · December 22nd 2006, 05:18 AM

Hello, m777!

1) The derivative of a continuous function is given.
Find all critical points and determine if a rel.max., rel.min. or neither occurs there.

$f'(x)\:=\:2\sin^3x - \sin^2x,\quad 0 < x <2\pi$

Solve $f'(x) = 0$
. . $2\sin^3x - \sin^2x \:= \:0\quad\Rightarrow\quad \sin^2x\left(2\sin x - 1\right)\:=\:0$

We have two equations to solve:
. . $\sin^2x \,= \,0\quad \Rightarrow\quad \sin x \,= \,0\quad \Rightarrow \quad \boxed{ x \:=\:\pi}$
. . $2\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{1}{2} \quad\Rightarrow\quad \boxed{ x \:=\:\frac{\pi}{6},\,\frac{5\pi}{6}}$

Second derivative: . $f''(x) \:=\:6\sin^2x\cos x - 2\sin x\cos x\:=\:2\sin x\cos x\left(3\sin x - 1\right)$

$f''(\pi) \:=\:2\!\cdot\!\sin\pi\!\cdot\!\cos\pi(3\sin\pi - 1) \:=\:2\cdot0\cdot(\text{-}1)\,[3\cdot0 - 1]$ $\:=\:0\quad\hdots\quad\boxed{\text{ neither at }x = \pi}$

$f''\left(\frac{\pi}{6}\right) \:=\:2\!\cdot\!\sin\frac{\pi}{6}\!\cdot\cos\frac{\ pi}{6}\left(3\sin\frac{\pi}{6} - 1\right) \:=\:2\!\left(\frac{1}{2}\right)\!\left(\frac{\sqr t{3}}{2}\right)\!\left(3\!\cdot\!\frac{1}{2} - 1\right)\:>\:0$
. . $f(x)$ is concave up: $\cup\quad\hdots\quad\boxed{\text{relative minimum at }x = \frac{\pi}{6}}$

$f''(x)\left(\frac{5\pi}{6}\right) \:=\:2\cdot\sin\!\frac{5\pi}{6}\!\cdot\cos\!\frac{ 5\pi}{6}\left(3\sin\!\frac{5\pi}{6} - 1\right)$ $\:=\:2\!\left(\frac{1}{2}\right)\!\left(\text{-}\frac{\sqrt{3}}{2}\right)\!\left(3\!\cdot\!\frac{ 1}{2} - 1\right) \:<\:0$
. . $f(x)$ is concave down: $\cap\quad\hdots\quad\boxed{\text{relative maximim at }x = \frac{5\pi}{6}}$

**earboth** · December 22nd 2006, 10:30 AM

#3 Integrate it with respect to “y”

(i) $\int (2+y^2)^2dy$

Integrate it with respect to “x”
(ii) $\int \frac{1}{x^6} dx$

Hello,

to (i). Expand the bracket first and then integrate:

$\int (2+y^2)^2dy=\int(4+4y^2+y^4)dy=4y+\frac{4}{3}y^3+\ frac{1}{5}y^5+C$

to (ii):

$\int \frac{1}{x^6} dx=\int x^{-6}dx=-\frac{1}{5}x^{-5}+C$

EB

Merry Christmas and a happy New Year!

**m777** · December 22nd 2006, 10:54 AM

Hello,
Merry christmas and happy new year to all of you.specially to soroban, earboth and captain black.
from
m777

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once again thanks.

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