Final answer:
To linearize the function f(x) = √(8x) at a = 1, the derivative is calculated and used to form the equation of the tangent line. The correct linearization is L(x) = 2√2 + √2(x - 1), and using it to approximate f(9.02) gives about 27.12, which agrees with option B.
Step-by-step explanation:
To find the linearization of the function f(x) = √(8x) at a = 1, we first calculate the derivative of f(x) evaluated at x=1. This gives us the slope of the tangent line at that point, which we will use to create the linear approximation.
The derivative of f(x) with respect to x is f'(x) = 4/√(8x). Evaluating this at x=1 gives f'(1) = 4/√(8), which simplifies to 4/2√2 = 2/√2. Rationalizing the denominator, we get 2√2/2 = √2.
Hence, the linearization of the function at a=1 is L(x) = f(1) + f'(1)(x - 1). Since f(1) = √8 = 2√2 and we previously found f'(1) = √2, the linearization becomes L(x) = 2√2 + √2(x - 1).
To approximate f(9.02), we substitute x = 9.02 into our linear equation: L(9.02) = 2√2 + √2(9.02 - 1). After performing the calculations, we find that the approximation is f(9.02) ≈27.12, which matches option B.