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- anonymous

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- anonymous

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- anonymous

Does that say \( \large f(x) = 4x + e^2x \)

- anonymous

4x+e^2x

- anonymous

e^(2x)

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- anonymous

Yes

- mathmate

@camila1315
or is it:
\(\large f(x) = 4x + e^{2x}\)

- anonymous

Yes thats it

- 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

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

- 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

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

*linear approximation

- 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

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

my result came to be between 5% and 10%
I can show you my reasoning again

- 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

\[ \large \Delta y \approx \Delta x \left[ \frac{dy}{dx} \right]_{x=a}
\\ ~ \\ \large \sf \% error = \left |\frac{\Delta y - \Delta x \left[ \frac{dy}{dx} \right]_{x=a} }{\Delta y }\right| \cdot 100
\]

- anonymous

Yeah I think math is right

- anonymous

let me know if you get it correct. I don't think its right.

- anonymous

Okay I will. Thank you both for all the help!!!

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