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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does that say \( \large f(x) = 4x + e^2x \)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1@camila1315 or is it: \(\large f(x) = 4x + e^{2x}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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) (xa)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The actual error \( \Delta y \) can be approximated as follows $$ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} $$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0delta 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%

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I 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)+(1.11.0)*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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1*linear approximation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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%

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my result came to be between 5% and 10% I can show you my reasoning again

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} \\ ~ \\ \large \sf \% error = \left \frac{\Delta y  \Delta x \left[ \frac{dy}{dx} \right]_{x=a} }{\Delta y }\right \cdot 100 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I think math is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me know if you get it correct. I don't think its right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I will. Thank you both for all the help!!!
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