Answer. We have a function f(x) = e^(2x+1). It is differentiable at point a=0, so for all x that are sufficiently close to a we have thefollowing approximation of f by linear function:
f(x) ~ f(a) + f'(a) * x In our case f(a) = e^(2*0+1) = e f'(x) = 2*e^(2x+1) f'(a) = 2e^(2*0+1) = 2e. Therefore the linearization of f at a=0 has the followingform: f(x) ~ e + 2ex.
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