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Find the linearization of f(x)=√8xa=1 and use it to approximate f(9.02).

A) Linearization: L(x)=3+3(x−1), Approximation: f(9.02)≈27.06
B) Linearization: L(x)=3+6(x−1), Approximation: f(9.02)≈27.12
C) Linearization: L(x)=3+2(x−1), Approximation: f(9.02)≈26.04
D) Linearization: L(x)=3+4(x−1), Approximation: f(9.02)≈26.08

User Racer
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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.

User Takien
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