Answer:
L(x) = 1 + nx is the linearization of the function f(x) = (1 + nx)^n at 0.
Explanation:
To show that L(x) = 1 + nx is the linearization of the function
f(x) = (1 + nx)^n at 0, we need to verify two conditions: the value of L(x) at x = 0 should be equal to f(0), and the derivative of L(x) should be equal to the derivative of f(x) at x = 0.
Evaluating L(x) at x = 0: L(0) = 1 + n(0) = 1
Evaluating f(0): f(0) = (1 + n(0))^n = 1^n = 1
Since L(0) = f(0), the first condition is satisfied.
Taking the derivative of L(x): L'(x) = n
Taking the derivative of f(x): f'(x) = n(1 + nx)^(n-1)
Evaluating f'(0): f'(0) = n(1 + n(0))^(n-1) = n(1)^n = n
Since L'(x) = f'(0), the second condition is satisfied.
Therefore,
We have shown that L(x) = 1 + nx is the linearization of the function
f(x) = (1 + nx)^n at 0.