For f(x)=exf(x) = e^xf(x)=ex near x=0x = 0x=0, the tangent-line approximation L(x)L(x)L(x) equalsA111BxxxC1+x1 + x1+xcheck_circleDe⋅xe \cdot xe⋅xExplanationf(0)=1f(0) = 1f(0)=1, f′(0)=1f'(0) = 1f′(0)=1. L(x)=1+xL(x) = 1 + xL(x)=1+x.