Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 10

$-0.763$ and $0.6072$

$e^xcosx \approx 1 +x$ $f(x) = e^xcosx$ $f(0) = e^0 cos(0)$ $f(0) = 1$ $f'(x) = e^xcosx + e^x(-sinx)$ $f'(x) = e^xcosx - e^xsinx$ $f'(0) = e^0cos(0) - e^0sin(0)$ $f'(0) = 1$ $L(x) = f(0) + x(f'(0))$ $L(x) = 1 + x(1)$ $L(x) = 1 + x$ $L(x) = 1 + x(1)$ This means that the original answer given by the book is correct. Now we have to find the values of x for which the linear approximation is accurate to within 0.1. $f(x) - L(x) < 0,1$ $e^xcosx -x -1 < 0.1$ $-0.1 < e^xcosx -x -1 < 0.1$ $-0.1 + 1 < e^xcosx -x < 0.1 + 1$ $0.9 < e^xcosx -x < 1.1$ Using the graph of $e^xcosx -x$ and the values of $y$ we can see the approximations to be $-0.763$ and $0.6072$