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Math Help - Linearization problems... very confused:(

  1. #1
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    Exclamation Linearization problems... very confused:(

    Here are some problems I have in a review test tomorrow.

    Which of the following is an equation of the line tangent to the graph of f(x) = x^6 - x^4 at the point where f'(x) = -1

    Consider the function f(x) -sin(kx) + 3 Given that F' (0) = k, what is the approximate value of f(0.03)

    Consider the function f(x) = aln(x+2). Given that f'(1)= a/3 what is the approximate value of f(0.98)

    I know for easier problems you use y - y1 = m(x-x1)
    but for these problems you are missing x i think.. right?
    I am so confused
    I would really appreciate it if someone guided me the right way... thank you!
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  2. #2
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    In the future, post at most two problems per thread, and then it's preferred if both problems are in the OP, and not added later.

    For all these problems, you're being asked to first find the equation of a line, for which you need a point on the line, and its slope. Then, you're asked to find another point on that line.

    Concerning your first problem: you need to find the x at which the derivative is -1. So, what do you get for f'(x)?

    Concerning your second problem: you already have x = 0 for the point of linearization. All you need here is f(0), which is what?

    Concerning your third problem: I have no idea what the aln function is. You're theoretically given the same information as in your second problem. All you really need is f(1), which is what?
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  3. #3
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    ahh..sorry..
    honestly i am not understanding :S
    for 1 i wrote: y- (x^6-x^4) = -1 (x-x1)
    i feel like im over thinking ....
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  4. #4
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    You're not thinking clearly, is the problem. As I said: in order to find the equation of a line, you need a point on that line, and you need the slope. The slope you have, but you don't have the point. You don't have either the x or y coordinates of the point. But you do have the equation of the function itself. How could you find the x-coordinate of a point on the line?
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  5. #5
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    mm is it y= -x - .836?
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    Show me your steps. How did you get that?
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  7. #7
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    well the equation of al ine is y=mx+b
    so i put
    y=(-1)x + b
    y= -x + b...
    then i used my graphing calculator and clicked the point where the tangent line would be and it gave me -.836
    im not sure how to find b without a calculator...

    unless i put f(x) = -x + b.... which would be
    x^6-x^4 = -x + b ..... but that doesn't really make sense and im confusing myself even more T.T
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  8. #8
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    That procedure is not going to be very accurate. In addition, it does not increase understanding whatsoever.

    The only information you're given about the point about which you are finding the tangent line, is the function's expression, and the value of the slope. Why not take the derivative of the function? What could you do with that?
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  9. #9
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    f(x)= x^6-x^4
    f'(x)= 6x^5-4x^3
    would i then plug in -1? or set it equal to -1?
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  10. #10
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    would i then plug in -1? or set it equal to -1?
    You tell me! Which one makes more sense? What kind of a number is the -1 given to you?
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  11. #11
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    you set it = to -1.......
    then i factored x^3 out....
    -1 = x^3 (6x^2 - 4)
    -1= 6x^2 - 4
    6x^2=5
    x = sqrt ( 5/6 )
    which = .913?
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  12. #12
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    You may not factor the x^3 out, because the product is not equal to zero but -1. Think to yourself: if a product is equal to -1, does that imply that either of the multiplicands is equal to -1? You could have ab = -1, with a = 5, and b = -1/5. The product is -1, but neither multiplicand is equal to -1.

    I'm afraid you're going to have to find the root numerically. You have yourself a non-factoring quintic polynomial to solve. The general quintic is not solvable by any exact method, so I would advise a numerical technique. See here for one answer. You can see that WolframAlpha gave one real answer, so I would go with that.
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  13. #13
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    thank you! these problems better not be on the quiz because my teacher said it's a noncalculator test :O
    thanks for your help...
    now would i use the same steps to solve the other two?
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  14. #14
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    Same idea, but the other two problems are actually easier, because you're already given the x-coordinate of the point on the tangent line. All you have to do, basically, is find the y-coordinate for the point, and then plug into the usual y = mx + b equation.
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  15. #15
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    THANK YOU SO MUCH. I feel like I have given you a hard time! I'm sorry
    I've been absent for the past two days because I'm sick and I have a test tomorrow and I am trying to learn the new material..
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