Final answer:
The local linearization of f(x) = e^(x^5) near x = 1 is y = 5e(x - 1) + e.
Step-by-step explanation:
The local linearization of f(x) = e^(x^5) near x = 1 is:
The equation of the tangent line at x = 1 is y = f'(1)(x - 1) + f(1), where f'(1) is the derivative of f(x) evaluated at x = 1, and f(1) is the value of f(x) at x = 1.
To find the derivative, we use chain rule. Let u = x^5, then f(x) = e^u. The derivative is f'(x) = e^u * u', where u' is the derivative of u with respect to x.
Using the power rule, we have u' = 5x^4. Substituting back, f'(x) = e^(x^5) * 5x^4.
Evaluating f'(1) and f(1), we get f'(1) = 5e and f(1) = e.
Therefore, the local linearization at x = 1 is y = 5e(x - 1) + e.