Local Linearization: Finding the Formula of a Graph

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The discussion revolves around finding the correct formula for a graph using local linearization, specifically at the point (3, 2). One participant initially derived the equation y = 2x + 4, while the book's answer was 2x - 8, leading to confusion about the correct approach. Clarification was sought regarding the point through which the line should pass, with a suggestion to use the point (3, 2) instead of (-2, 0). The conversation also touched on the importance of understanding the geometric meaning of local linearization and encouraged working through simpler problems for clarity. Ultimately, the correct local linearization method was confirmed, emphasizing the need for careful selection of points in the process.
UrbanXrisis
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the question is http://home.earthlink.net/~urban-xrisis/clip001.jpg

I got a different answer than what the book says...

so I need to find the formula of the graph.

H'(3)=f(3)=2
m=\frac{\Delta y}{\Delta x}

2=\frac{\Delta y}{x- \int^3_0 f(t)dt}
y=2(x+2)
y=2x+4

the book's answer is 2x-8

where did I go wrong?
 
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Through what point did you want your line to go through? It looks like you used (-2, 0)...
 
you mean I should do:
2=\frac{y-2}{x-3}
y=2x-4
??

what I did was...

H'(x)=\frac{y-\int_{-2}^yf(t)dt}{x-\int_0^xf(t)dt}
 
Why do you want your line to go through the point (3, 2)?

What you need to do is stop guessing and think it through. Working through a simpler problem might help.

What is the local linearization of the function f(x) = x2 near x = -1? First tell me what that means geometrically, then work out the answer algebraically.
 
f'(x)= (y2-y1) / (x2-x1)
-2= (y2-1) / (x+1)
y=-2x-1
 
thank you, I used your example to get the right answer
 
I notice you didn't try a geometric explanation. :-p

Anyways, that's exactly right. Now, why did you pick the point (x1, y1) = (-1, 1)? Apply the same reasoning to your problem.
 

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