Answer to Question #26777 in Calculus for alex
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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