Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 27

$e^{0.1} \approx 1.1$

$f(a) = e^a$ $f(0) = e^0$ $f(0) = 1$ $f'(x) = e^a$ $f'(0) = e^0$ $f'(0) = 1$ $L(x) = f(a) + f'(a)(x-a)$ $L(x) = 1 + 1(x-0)$ $L(x) = 1 + x - 0$ $L(x) = 1 + x$ $L(0.1) = 1 + 0.1$ $L(0.1) = 1.1$ $e^{0.1} \approx 1.1$