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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Does that say \( \large f(x) = 4x + e^2x \)

  2. anonymous
    • one year ago
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    4x+e^2x

  3. anonymous
    • one year ago
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    e^(2x)

  4. anonymous
    • one year ago
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    Yes

  5. mathmate
    • one year ago
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    @camila1315 or is it: \(\large f(x) = 4x + e^{2x}\)

  6. anonymous
    • one year ago
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    Yes thats it

  7. anonymous
    • one year ago
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    The local linear approximation at x = a is essentially the tangent line approximation at the point. If x ≈ a , then f(x) ≈ f(a) + f '(a) (x-a)

  8. anonymous
    • one year ago
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    The actual error \( \Delta y \) can be approximated as follows $$ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} $$

  9. anonymous
    • one year ago
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    delta x here is 1.1 - 1 = .1 dy/dx = 4 + e^(2x) * 2 plug in x = 1 estimated delta y = .1 ( 4 + 2*e^(2*1))=1.8778 actual delta y is f(1.1) - f(1) = 2.035957 the percent error of this approximation is ( 2.035957 - 1.8778) / (2.035957) * 100 = 7.77%

  10. mathmate
    • one year ago
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    I suggest to compare the approximate and accurate values of f(1.1). With \(\large f(x) = 4x + e^{2x}\) \(\large f'(x)=2e^{2x}+4\) f(1.0)=11.3891 exact=f(1.1)=13.4250 approx=f(1.0)+(1.1-1.0)*f'(1.0)=13.2669 approx/exact=13.2669/13.4250=0.9882 i.e. the approximate is short by 1.18% |dw:1440283809059:dw| when we examine the curve of f(x) above, it is obvious that the linear approximate understimates the value of f(1.1) because the slope is constantly increasing.

  11. mathmate
    • one year ago
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    *linear approximation

  12. anonymous
    • one year ago
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    thats odd we get two different answers your reasoning looks valid, but so does mine. i will be back later and try to figure out the discrepancy

  13. anonymous
    • one year ago
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    I have answer choices if you have different answers A. Between 0% and 4% B. Between 5% and 10% C. Between 11% and 15% D. Greater than 15%

  14. anonymous
    • one year ago
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    my result came to be between 5% and 10% I can show you my reasoning again

  15. mathmate
    • one year ago
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    I don't think it is a difference of the math, it's a difference in the interpretation of the question: " the % error of \(this\) approximation is" I interpret it as asking for the error of f(1.1), and you interpret it as asking for the error of f(1.1)-f(1).

  16. anonymous
    • one year ago
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    \[ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} \\ ~ \\ \large \sf \% error = \left |\frac{\Delta y - \Delta x \left[ \frac{dy}{dx} \right]_{x=a} }{\Delta y }\right| \cdot 100 \]

  17. anonymous
    • one year ago
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    Yeah I think math is right

  18. anonymous
    • one year ago
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    let me know if you get it correct. I don't think its right.

  19. anonymous
    • one year ago
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    Okay I will. Thank you both for all the help!!!

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