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Linear approximations

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    explain in terms of linear approximations why the approximation is reasonable.

    2. Relevant equations
    L(x)=f(a) + f'(a)(x-a)

    3. 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

    Is that the right equation?
    Because in class the equation was
    which is the correct answer. But why is y1=1 and not 6??? Can anyone catch that?
  2. jcsd
  3. Oct 3, 2007 #2

    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...

  4. Oct 3, 2007 #3
    They didn't give me P(1,1) though
    I arbitrarily picked x=1 to solve the original f'(x) equation.
  5. Oct 3, 2007 #4
    Isn't my point (1,6)
    I plugged in x=1 and got f'(x)=6
  6. Oct 3, 2007 #5
    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)?
  7. Oct 3, 2007 #6
    So f'(x)=6, but f(x)=1 so f(x) is the value I should put in for y in (x1,y1)?
  8. Oct 4, 2007 #7


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    Science Advisor

    I hope it was NOT "arbitrary". The fact that 1 is very close to 1.01 should have guided you!

    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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