anonymous
  • anonymous
....
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Does that say \( \large f(x) = 4x + e^2x \)
anonymous
  • anonymous
4x+e^2x
anonymous
  • anonymous
e^(2x)

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anonymous
  • anonymous
Yes
mathmate
  • mathmate
@camila1315 or is it: \(\large f(x) = 4x + e^{2x}\)
anonymous
  • anonymous
Yes thats it
anonymous
  • anonymous
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)
anonymous
  • anonymous
The actual error \( \Delta y \) can be approximated as follows $$ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} $$
anonymous
  • anonymous
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%
mathmate
  • mathmate
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.
mathmate
  • mathmate
*linear approximation
anonymous
  • anonymous
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
anonymous
  • anonymous
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%
anonymous
  • anonymous
my result came to be between 5% and 10% I can show you my reasoning again
mathmate
  • mathmate
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).
anonymous
  • anonymous
\[ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a} \\ ~ \\ \large \sf \% error = \left |\frac{\Delta y - \Delta x \left[ \frac{dy}{dx} \right]_{x=a} }{\Delta y }\right| \cdot 100 \]
anonymous
  • anonymous
Yeah I think math is right
anonymous
  • anonymous
let me know if you get it correct. I don't think its right.
anonymous
  • anonymous
Okay I will. Thank you both for all the help!!!

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