Introduction:
The problem requires finding the value of alpha in the function f(x) = √x * alpha x, given that g(x) is the sum of the first three terms of the Taylor series of f(x) around x=1 and g(3) = 3.

Approach:
To solve this problem, we need to find the Taylor series expansion of f(x) around x=1 and then evaluate it at x=3. Since g(x) is the sum of the first three terms of the Taylor series, we can equate g(x) to the Taylor series expansion of f(x) and substitute x=3 to find the value of alpha.

Step-by-Step Solution:

1. Find the Taylor series expansion of f(x) around x=1:
The Taylor series expansion of f(x) around x=1 can be found by taking derivatives of f(x) with respect to x and evaluating them at x=1. Since f(x) = √x * alpha x, we can write it as:

f(x) = alpha * √x * x

2. Find the derivatives of f(x):
To find the derivatives of f(x), we need to apply the product rule and chain rule.

f'(x) = alpha * (√x * x)' = alpha * (0.5 * x^(-0.5) * x + √x) = alpha * (0.5 * √x + √x) = alpha * 1.5 * √x

f''(x) = alpha * (1.5 * √x)' = alpha * (1.5 * 0.5 * x^(-0.5)) = alpha * 0.75 * x^(-0.5) = alpha * 0.75 / √x

f'''(x) = alpha * (0.75 / √x)' = alpha * (-0.75 * 0.5 * x^(-1.5)) = alpha * (-0.375 * x^(-1.5)) = alpha * (-0.375) / (x^(1.5))

3. Write the Taylor series expansion of f(x) around x=1:
The Taylor series expansion of f(x) around x=1 can be written as:

f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2! + f'''(1)(x-1)^3/3! + ...

Since we only need the first three terms, we can write:

f(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2!

4. Evaluate the Taylor series at x=3:
We are given that g(x) = a0 + a1(x-1) + a2(x-1)^2 is the sum of the first three terms of the Taylor series of f(x) around x=1. Therefore, we can equate g(x) to the Taylor series expansion of f(x) and substitute x=3:

g(x) = f(1) + f'(1)(x-1) + f''