Newtons method and distance help

[camboguy]

Homework Statement

Given: Let d be the distance function of a point on a parabola y=9+3x-x^2 and point (5,11)

Questions:
a) express f(x) = d^2 in terms of x
b) show that there is only one critical point of f
c) approximate the critical point by Newton's method with the initial guess x sub zero = 2.5 (accurate to 3 decimal places). Is this a local maximum, local minimum or neither?
d) find the furthest and the closest pings on the parabola to the ping (5,11) where 0 <= x <= 5

Homework Equations

The Attempt at a Solution

i get a,b,and part of c. but I am stuck in part c where they ask me if the point is a local maximum,minimum,or neither.

part c what i did/need help) i did Newtons method with x sub n -f'(x)/f''(x)
and got that the zero is about 2.635999161. and I am confused on how to find if it is a max,min, or neither.

part d what i need did/ need help) i am totally stuck on this one i have no clue where to start it. i drew a few pictures and tried a few thing but it always got me to a dead end. could you help me on this too.



Thanks Before hand.

 

[Mark44]

I found f(x) = d^2 = x^4 - 6x^3 + 14x^2 -22x + 29,
from which I found f'(x) = 4x^3 - 18x^2 + 28x - 22

I wasn't able to factor the cubic, so I wasn't able to find critical points using the derivative.

It's clear geometrically, though, that there is just one critical point. By graphing the parabola, I see that it passes through (0, 9) and (3, 9) and that the vertex of the parabola (which opens downward) is at (3/2, 45/4). The point (5, 11) is to the right of the parabola and at a lower level than the vertex so there will be one point on the parabola that is closest, hence f'(x) = 0 for just one value. The critical point will be for the minimum distance. We can make f(x) arbitrarily large by taking larger or more negative values of x.

I'm curious as to what equation you ran Newton's method on. I hope you ran it on the equation f'(x) = 0, since it's clear to me that the equation f(x) = 0 doesn't have any solutions.

 

[Mark44]

Okay, so on part C I'm pretty sure you messed up when using the formula.
y=9+3x-x²
y'=-2x+3
y''=-2
Now, you had the right idea using n_n - y'/y'', and 2.5 as an initial guess would tell you that the answer is x=1.5 or 3/2. this tell us that we have a maximum value at x=1.5 because the second derivative is negative.

and as for part D, I'm not sure what a ping is.

By the way. Look at a graph of the equation. It can help a lot.

Take another look at the problem the OP posted. It was to find the minimum distance between the parabola and the point (5, 11), not to find the high point on the parabola.

 

[lucidicblur]

Take another look at the problem the OP posted. It was to find the minimum distance between the parabola and the point (5, 11), not to find the high point on the parabola.

Yeah, I just saw that. My mistake. What is a ping though?

 

[Mark44]

I think he might have meant point, but not sure. Otherwise, I don't have any idea.

 

[HallsofIvy]

Perhaps he is getting a sonar picture of the parabola!

 

[camboguy]

opps that should say point.