Why is the Linear Approximation Reasonable for (1.01)^6?

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Homework Help Overview

The discussion revolves around the use of linear approximation to estimate the value of (1.01)^6. Participants are exploring the reasoning behind the approximation and the correct application of the linear approximation formula.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the choice of points for linear approximation, specifically questioning why the point (1,1) is used instead of (1,6). There is an exploration of the relationship between the function and its derivative, as well as the implications of selecting different points for approximation.

Discussion Status

The discussion is active, with participants questioning assumptions about the function values and derivatives. Some guidance has been offered regarding the significance of the chosen point in relation to the approximation, but no consensus has been reached on the interpretation of the linear approximation setup.

Contextual Notes

Participants express uncertainty about the appropriateness of their chosen points for linear approximation and the implications of the derivative's value at those points. There is a recognition that the proximity of 1 to 1.01 is relevant to the approximation process.

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Homework Statement


explain in terms of linear approximations why the approximation is reasonable.
(1.01)^6=1.06



Homework Equations


y-y1=m(x-x1)
L(x)=f(a) + f'(a)(x-a)


The Attempt at a Solution


given 1.01^6, f(x)=x^6, so f'(x)=6x^5
plugging in x=1, f'(1)=6
y-6=6(x-1)

Is that the right equation?
Because in class the equation was
y-1=6(x-1)
y=1.06
which is the correct answer. But why is y1=1 and not 6? Can anyone catch that?
 
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fk378 said:
which is the correct answer. But why is y1=1 and not 6? Can anyone catch that?

y-y1=m(x-x1)

Where m is your slope and the point given is (x1,y1), since f(1)=1 your point is (1,1)
f'(1)=6=m so...

y-1=6(x-1)
 
They didn't give me P(1,1) though
I arbitrarily picked x=1 to solve the original f'(x) equation.
 
Feldoh said:
y-y1=m(x-x1)

Where m is your slope and the point given is (x1,y1), since f(1)=1 your point is (1,1)
f'(1)=6=m so...

y-1=6(x-1)

Isn't my point (1,6)
I plugged in x=1 and got f'(x)=6
 
fk378 said:
Isn't my point (1,6)
I plugged in x=1 and got f'(x)=6

Yes f'(1)=6 but that doesn't mean when x=1 f(x)=6 necessarily. The point of a function f(x) at x is (x,f(x)). What does the derivative tell us about f(x)?
 
So f'(x)=6, but f(x)=1 so f(x) is the value I should put in for y in (x1,y1)?
 
fk378 said:
They didn't give me P(1,1) though
I arbitrarily picked x=1 to solve the original f'(x) equation.
I hope it was NOT "arbitrary". The fact that 1 is very close to 1.01 should have guided you!

fk378 said:
So f'(x)=6, but f(x)=1 so f(x) is the value I should put in for y in (x1,y1)?
f(x) is NOT "1", the function you are looking at is f(x)= x6. What is f(1)= (1)6?
 

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