Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a.
ƒ(x) = sinh (-3x), n = 3, a = 0

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Recall the definition of the nth-order Taylor polynomial of a function \( f(x) \) centered at \( a \): \[ T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k \] where \( f^{(k)}(a) \) is the \( k \)-th derivative of \( f \) evaluated at \( a \).

Identify the function and the center: here, \( f(x) = \sinh(-3x) \), \( n = 3 \), and \( a = 0 \). We will need to find \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f^{(3)}(0) \).

Compute the derivatives of \( f(x) = \sinh(-3x) \) step-by-step: - First derivative: \( f'(x) = \frac{d}{dx} \sinh(-3x) \) - Second derivative: \( f''(x) = \frac{d}{dx} f'(x) \) - Third derivative: \( f^{(3)}(x) = \frac{d}{dx} f''(x) \) Remember to apply the chain rule carefully since the argument is \( -3x \).

Evaluate each derivative at \( x = 0 \): Calculate \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f^{(3)}(0) \) by substituting \( x = 0 \) into each derivative expression.

Construct the Taylor polynomial of order 3 using the formula: \[ T_3(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 \] This polynomial approximates \( f(x) \) near \( x = 0 \).

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