# Answer to Question #26777 in Calculus for alex

Question #26777

linearization of e^(2x+1) at a=0

Expert's answer

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.

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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