Answer
$L\left( x \right) = 3x + 1$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^{3x}},{\text{ }}a = 0 \cr
& {\text{Evaluate }}f\left( x \right){\text{ at }}a = 0 \cr
& f\left( 0 \right) = {e^{3\left( 0 \right)}} \cr
& f\left( 0 \right) = 1 \cr
& {\text{Differentiating the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{3x}}} \right] \cr
& f'\left( x \right) = 3{e^{3x}} \cr
& {\text{Evaluate }}f'\left( x \right){\text{ at }}a = 0 \cr
& f'\left( 0 \right) = 3{e^{3\left( 0 \right)}} \cr
& f'\left( 0 \right) = 3 \cr
& {\text{The linear approximation at }}f\left( a \right){\text{ is given by: }}\left( {{}} \right){\text{ }} \cr
& L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( 0 \right) = 1,{\text{ }}f'\left( 0 \right) = 3{\text{ and }}a = 0{\text{ into eq.}}\left( {\bf{1}} \right) \cr
& L\left( x \right) = 1 + 3\left( {x - 0} \right) \cr
& {\text{Simplifying}} \cr
& L\left( x \right) = 1 + 3x \cr
& or \cr
& L\left( x \right) = 3x + 1 \cr} $$
