Answer
$$a = 0,{\text{ }}{e^{0.06}} \approx 1.06$$
Work Step by Step
$$\eqalign{
& {e^{0.06}} \cr
& {\text{Rewrite }}{e^{0.06}}{\text{ as }}{e^{0 + 0.06}} \cr
& {\text{Using the function }}f\left( x \right) = {e^x},{\text{ }} \cr
& f\left( x \right) = {e^x} \cr
& {\text{we have that }}0.06 = 0 + 0.06,{\text{ let }}a = 0 \cr
& {\text{* Evaluate }}f\left( 0 \right) \cr
& f\left( 0 \right) = {e^0} \cr
& f\left( 0 \right) = 1 \cr
& \cr
& {\text{Differentiating we obtain}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}} \right] \cr
& f'\left( x \right) = {e^x} \cr
& {\text{* Evaluate }}f'\left( 0 \right) \cr
& f'\left( 0 \right) = {e^0} \cr
& f'\left( 0 \right) = 1 \cr
& \cr
& {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }}\left( {page{\text{ 282}}} \right) \cr
& L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr
& \cr
& {\text{The linear approximation }}L\left( x \right){\text{ at }}a = 0{\text{ is:}} \cr
& L\left( x \right) = 1 + 1\left( {x - 0} \right) \cr
& L\left( x \right) = x + 1 \cr
& \cr
& {\text{Estimating the given value at }}f\left( {0.06} \right) \cr
& L\left( x \right) = x + 1 \cr
& L\left( {0.06} \right) = 0.06 + 1 \cr
& L\left( {0.06} \right) = 1.06 \cr
& {\text{Therefore}}{\text{,}} \cr
& {e^{0.06}} \approx L\left( {0.06} \right) \approx 1.06 \cr} $$
