To find the degree-three Taylor polynomial $T_3(x)$ for $f(x) = \ln(2 - 3x)$ centered at $x = 0$, we follow these steps:

1. Compute the derivatives of $f(x)$:

We begin by calculating the first few derivatives of $f(x)$.

$f(x)$:

$f(x) = \ln(2 - 3x)$

First derivative $f'(x)$:

$f'(x) = \frac{d}{dx} \ln(2 - 3x) = \frac{-3}{2 - 3x}$

Second derivative $f''(x)$:

$f''(x) = \frac{d}{dx} \left(\frac{-3}{2 - 3x}\right) = \frac{-3 \cdot (-3)}{(2 - 3x)^2} = \frac{9}{(2 - 3x)^2}$

Third derivative $f'''(x)$:

$f'''(x) = \frac{d}{dx} \left(\frac{9}{(2 - 3x)^2}\right) = \frac{9 \cdot 2(-3)}{(2 - 3x)^3} = \frac{-54}{(2 - 3x)^3}$

Evaluate derivatives at $x = 0$:

At $x = 0$, the derivatives are:

• $f(0) = \ln(2)$
• $f'(0) = \frac{-3}{2}$
• $f''(0) = \frac{9}{4}$
• $f'''(0) = \frac{-54}{8} = -\frac{27}{4}$

2. Construct the Taylor polynomial:

The Taylor polynomial centered at $x = 0$ is given by: $T_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!}$

Substitute the values: $T_3(x) = \ln(2) + \left(-\frac{3}{2}\right)x + \frac{\frac{9}{4}x^2}{2} + \frac{-\frac{27}{4}x^3}{6}$

Simplify each term:

• The $x^2$-term: $\frac{\frac{9}{4}}{2} = \frac{9}{8}$
• The $x^3$-term: $\frac{-\frac{27}{4}}{6} = -\frac{27}{24} = -\frac{9}{8}$

Thus: $T_3(x) = \ln(2) - \frac{3}{2}x + \frac{9}{8}x^2 - \frac{9}{8}x^3$

Final Answer:

The degree-three Taylor polynomial is: $T_3(x) = \ln(2) - \frac{3}{2}x + \frac{9}{8}x^2 - \frac{9}{8}x^3$

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Would you like a step-by-step explanation of the derivation process or help with related concepts? Here are some related questions to consider:

1. How do you generalize Taylor series for a given function?
2. What are the practical applications of Taylor polynomials?
3. How do you check the accuracy of a Taylor polynomial approximation?
4. Can you explain how the radius of convergence impacts this Taylor series?
5. What happens if the Taylor polynomial is centered at a different point?

Tip: Always evaluate derivatives carefully at the center point for accurate coefficients in Taylor series!