# Dervatives and Newton's Method

• Apr 9th 2007, 04:15 PM
qbkr21
Dervatives and Newton's Method
#1. f(x) = x^4-17x^2+18

What is the relationship between the sign of f'(x) and the graph f(x) ?:eek:
What is the relationship between the roots of f''(x) and the graph of f(x) ?:eek:
What is the relationship between the sign of f''(x) and the graph of f(x) ?:eek:

#2. Given that f(x) = sin(x) use Newton's Method to find the first 2 positive zeros showing the first few approximations in each case.

So...

pi
___

x1: 3
x2: 3.142546543
x3: 3.141592653
x4: 3.14159255359

also...

2pi
____

x1: 6
x2: 6.29100619138
x3: 6.28310514772
x4: 6.28318530718

The first positive zero is: 3.14159
The second positive zero: 6.28319

What happens when you took pi/2 as first approximation? Explain

I got 1 but else is there to say:eek: :eek:

#3. if f(x) = x^4-17x^2+18
f''(x) = 12x^2-34

Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __:eek: ___ of f(x)?

Thanks for the Help!
• Apr 9th 2007, 04:34 PM
qbkr21
Re:
P.S. This is one of the Calculator labs that the department makes up
• Apr 9th 2007, 04:37 PM
Jhevon
Quote:

Originally Posted by qbkr21
#1. f(x) = x^4-17x^2+18

What is the relationship between the sign of f'(x) and the graph f(x) ?:eek:
What is the relationship between the roots of f''(x) and the graph of f(x) ?:eek:
What is the relationship between the sign of f''(x) and the graph of f(x) ?:eek:

the sign of f'(x) let's you know when f(x) is increasing or decreasing. if f'(x) is positive, f(x) has a positive slope and is hence increasing. if f'(x) is negative, f(x) has a negative slope and is hence decreasing

the roots of f''(x) give the inflection points of f(x), that is the points where f(x) changes concavity from concave up to concave down or vice versa.

the sign of f''(x) indicates the concavity (and hence the nature of critical points) of f(x). if f''(x) is positive for some critical point in f(x), then f(x) is concave up at that point and it is a local min. if f''(x) is negative at some critical point of f(x), then f(x) is concave down at that point and it is a local max

Quote:

Originally Posted by qbkr21

#3. if f(x) = x^4-17x^2+18
f''(x) = 12x^2-34

Therefore f''(x) = 0 if x = -sqrt(102)/6 and sqrt(102)/6 and these are the x-coordinates of the points of __:eek: ___ of f(x)?

points of inflection of f(x)...usually.
• Apr 9th 2007, 04:43 PM
qbkr21
Re:
Here is the exact question Jhevon by the way thanks for the help...

http://item.slide.com/r/1/46/i/2g0wi...ijaa4jCDYLtJ3/
• Apr 9th 2007, 04:48 PM
Jhevon
Quote:

Originally Posted by qbkr21
Here is the exact question Jhevon by the way thanks for the help...

http://item.slide.com/r/1/46/i/2g0wi...ijaa4jCDYLtJ3/

you used x_(n+1) = x_n - f(x_n)/f'(x_n) for each of these right? did you find the first few values for pi/2?
• Apr 9th 2007, 04:51 PM
qbkr21
Re:
Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts.
• Apr 9th 2007, 04:58 PM
Jhevon
Quote:

Originally Posted by qbkr21
Uhhh...I just plug in pi/2 into Newtons method formula and x_n-f(x)/f'(x) and got the number 1. Is there something else I should do. Thanks for the helpful posts.

ok so these questions are weird, i don't know exactly what answer they are looking for, but anyway, we can see a few patterns. note that Newton's method is used to approximate the roots of a function. for each approximation we begin with, using Newton's method puts us closer to the root that is closest to that value.

not that when we begin with x = 3, each new approximation is bigger than the last and they keep getting closer to pi, 3.1415... since pi is the closest root to 3

when we start with x = 6, each new approximation is again bigger than the last, that is because we are increasing to 2pi, the next closest root to 6

when we begin with pi/2, each new approximation begins to decrease, this is because we are moving towards the closest root, which is zero. i find this interesting though, since pi is as close to pi/2 as 0 is