No, that is not the answer.
Use a McLaurin series about x = 1
f(x) = e at x = 1
f'(x) = 2x*e^x^2 which = 2e at x = 1
f(x) = f(1) + f'(1)(x-1) + ...
= e + 2e(x-1)
What is the local linearization of e^x^2 near x=1?
Is the answer 1+2x
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