Final answer:
To find the linearization of the function f(x) = √(x+3) at a = 1, we first find the derivative of the function. Using the point-slope form, we construct the linearization equation y = (1/4)x + (7/4) and then use it to approximate the values √3.98 and √4.05.
Step-by-step explanation:
To find the linearization of the function f(x) = √(x+3) at a = 1, we first find the derivative of the function. The derivative is f'(x) = 1/(2√(x+3)). Next, we evaluate f'(1) to find the slope of the tangent line at x = 1. Substituting x = 1 into f'(x), we get f'(1) = 1/(2√4) = 1/4. Now, we use the point-slope form of a line to construct the linearization equation: y - f(1) = f'(1)(x-1). Plugging in f(1) = √4 and f'(1) = 1/4, we get y - 2 = (1/4)(x-1). Simplifying gives the linearization equation y = (1/4)x + (7/4).
To approximate √3.98 and √4.05 using the linearization, we substitute these values of x into the linear equation. Approximating √3.98, we have y = (1/4)(3.98) + (7/4) ≈ 1.995 + 1.75 ≈ 3.745. Approximating √4.05, we have y = (1/4)(4.05) + (7/4) ≈ 2.025 + 1.75 ≈ 3.775.