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Thread: Linearization question

  1. #1
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    Linearization question

    Find linearization of f (x)= x^(1/3) With center a = 27

    I got 1/27*x +2

    But then it asks: is this linearization greater than or smaller than f (x) when x is close to but not equal to 27?

    What does this question mean. I'm not sure what it is asking and how.
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  2. #2
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    Re: Linearization question

    let $f(x) = x^{1/3},~l(x)=\dfrac{x}{27}+2$

    let $\epsilon > 0$

    $f(27 +\epsilon) \overset{?}{\lessgtr} l(27 + \epsilon)$

    $f(27 -\epsilon) \overset{?}{\lessgtr} l(27 -\epsilon)$
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  3. #3
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    Re: Linearization question

    What it is asking is, "if x is close to 3, is x^{1/3} greater than or less than \left(\frac{1}{27}\right)x+ 2?" Equivalently, "is x^{1/3}- \left(\frac{1}{27}\right)x- 2 positive or negative?". Of course, one way of getting the "linearization" of x^{1/3} around x= 27 is to take the first two terms, (x- 27)^0 and [tex](x- 27)^1[/ex], of the Taylor's series about x= 3. What is the coefficient of the quadratic, (x- 27)^2 term? Is it positive or negative?
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  4. #4
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    Re: Linearization question

    Wait, i'm kinda confused. Does this involve summation/integral? Because we didn't learn that yet, and it was part of the review exam which my professor said wouldn't be on the actual exam (summation)
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  5. #5
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    Re: Linearization question

    But then it asks: is this linearization greater than or smaller than f (x) when x is close to but not equal to 27?

    What does this question mean. I'm not sure what it is asking and how.
    If a function is concave up at the point of tangency, then points on a tangent line close to the point of tangency will be below points on the function curve.

    If a function is concave down at the point of tangency, then points on the tangent line close to the point of tangency will be above points on the function curve.

    $f(x) = x^{1/3} \implies f'(x) = \dfrac{1}{3x^{2/3}} \implies f''(x) = -\dfrac{2}{9x^{5/3}}$

    note $f''(27) < 0 \implies f(x)$ is concave down, therefore the tangent line, $L(x)$, used to determine a linear approximation lies above the curve at values of $x$ close to $x=27$ ... $L(x) > f(x)$
    Attached Thumbnails Attached Thumbnails Linearization question-cubrt_tangent.jpg  
    Last edited by skeeter; Nov 10th 2017 at 09:25 AM.
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  6. #6
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    Re: Linearization question

    Thank YOU! This makes much more sense !!
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