Final answer:
The linearization of f at x = 0 is L(x) = 1 + x, the estimate of f at x = 0.1 is approximately 1.1, and the actual value of f at 0.1 is expected to be less than 1.1.
Step-by-step explanation:
Linear Approximation
Linear approximation is a method used to estimate the value of a function at a point close to where we know its value. In this case, we are given the function f with f(0) = 1 and f'(x) = cos(x^2). We'll carry out the following steps:
a) Find the Linearization of f at x = 0
The linearization L(x) of function f at x = 0 is given by,
L(x) = f(0) + f'(0)(x - 0)
Since f(0) = 1 and f'(0) = cos(0^2) = 1, the equation becomes:
L(x) = 1 + 1*(x - 0) = 1 + x
b) Estimate the Value of f at x = 0.1
We can now use the linearization to estimate f(0.1):
f(0.1) ≈ L(0.1) = 1 + 0.1 = 1.1
c) Actual Value of f at x = 0.1 - Greater or Lesser than Estimate?
Since the derivative f'(x) = cos(x^2) decreases as x increases, the function f is increasing at a decreasing rate. Therefore, the actual value of f(0.1) will likely be less than the linear approximation because the slope of the tangent line (derivative) overestimates f as x moves away from 0.
The actual value of f at x = 0.1 is likely less than the estimate 1.1 due to the decreasing rate of increase.