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To find the Maclaurin series of a function like \( f(x) = (2x-1)^2 \), we need to calculate the derivatives of the function at specific points. The derivatives help us understand how the function changes at different values of \( x \). Here's how it works:- **First Derivative**: Calculating the first derivative, \( f'(x) \), tells us the slope or rate of change of the function. For our function, \( f(x) \) becomes \( f'(x) = 4(2x-1) \). At \( x=0 \), \( f'(0) = -4 \). This means the slope of \( f(x) \) at \( x=0 \) is \(-4\).- **Second Derivative**: The second derivative, \( f''(x) \), shows us how the slope itself changes. When you differentiate \( f'(x) = 4(2x-1) \), you get \( f''(x) = 8 \). Evaluating at \( x = 0 \), \( f''(0) = 8 \), indicating the rate of change of \( f'(x) \).The derivatives are used for constructing a polynomial that approximates the function near \( x=0 \). In this case, it's crucial because without these derivatives, we couldn't find the correct terms in the Maclaurin series.

Polynomial expansion is a powerful technique in mathematics to express functions as infinite sums of terms. With a function like \( f(x) = (2x-1)^2 \), we look for a polynomial equivalent that approximates the function near a point (in Maclaurin's case, \( x=0 \)).- A polynomial is simpler to compute and evaluate compared to more complex functions. Thus, expanding a function into its polynomial form using derivatives can greatly simplify calculations. - Specifically, the Maclaurin series expansion provides a polynomial where the coefficients are derived from the derivatives of the function at \( x=0 \).For our example, the expansion becomes \( f(x) = 1 - 4x + 4x^2 + \cdots \). Each term is a building block from derivative evaluations which represent the function around \( x=0 \). This method of polynomial expansion is not just limited to basic functions but can apply to a wide array of functional forms.

Function evaluation at a specific point is a key part of constructing the Maclaurin series. To begin expanding the function \( f(x) = (2x-1)^2 \), we first find its value at \( x=0 \).- **Why \( x=0 \)?** In Maclaurin series, we expand around \( x=0 \) because it simplifies calculations and still provides excellent approximations near this center, making these series widely applicable in many scenarios.- We compute \( f(0) \) directly: for our function, this results in \( (-1)^2 = 1 \). The value tells us the first term of our series.By evaluating the original function and its successive derivatives at \( x=0 \), we extract each term of the Maclaurin series. These evaluations are foundational, as they ensure that the polynomial closely mimics or mirrors the actual function's behavior near \( x=0 \). Understanding function evaluation at \( x=0 \) is essential for successful series construction, as it sets the stage for how the polynomial terms line up with the function's behavior.