The problem asks to find the coefficients $c_0, c_1, c_2, c_3$ of the Taylor series for the function $f(x) = 9^x$, centered at $x = 0$.

Step 1: General Formula for the Taylor Series

The Taylor series of a function $f(x)$ centered at $x = 0$ is given by:

$f(x) = \sum_{n=0}^{\infty} c_n x^n$ where $c_n$ is the $n$-th coefficient and is found by:

$c_n = \frac{f^{(n)}(0)}{n!}$

Here, $f^{(n)}(0)$ represents the $n$-th derivative of $f(x)$ evaluated at $x = 0$.

Step 2: Function and Derivatives of $f(x) = 9^x$

We begin by calculating the function and its derivatives at $x = 0$.

1. $f(x) = 9^x$
2. First derivative: $f'(x) = \ln(9) \cdot 9^x$
3. Second derivative: $f''(x) = (\ln(9))^2 \cdot 9^x$
4. Third derivative: $f^{(3)}(x) = (\ln(9))^3 \cdot 9^x$

Step 3: Evaluate Derivatives at $x = 0$

At $x = 0$, $9^0 = 1$, so the function and its derivatives become:

1. $f(0) = 1$
2. $f'(0) = \ln(9)$
3. $f''(0) = (\ln(9))^2$
4. $f^{(3)}(0) = (\ln(9))^3$

Step 4: Compute the Coefficients

Now, use the formula $c_n = \frac{f^{(n)}(0)}{n!}$ to find each coefficient:

• $c_0 = \frac{f(0)}{0!} = \frac{1}{1} = 1$
• $c_1 = \frac{f'(0)}{1!} = \frac{\ln(9)}{1} = \ln(9)$
• $c_2 = \frac{f''(0)}{2!} = \frac{(\ln(9))^2}{2}$
• $c_3 = \frac{f^{(3)}(0)}{3!} = \frac{(\ln(9))^3}{6}$

Final Answers:

• $c_0 = 1$
• $c_1 = \ln(9)$
• $c_2 = \frac{(\ln(9))^2}{2}$
• $c_3 = \frac{(\ln(9))^3}{6}$

Would you like more details on Taylor series, or do you have any other questions? Here are 5 related questions to explore:

1. How is the radius of convergence of a Taylor series determined?
2. How can you approximate functions using Taylor series?
3. What is the importance of Taylor series in numerical analysis?
4. How does the Taylor series change if the function is centered at a point other than $x = 0$?
5. What are some applications of Taylor series in physics or engineering?

Tip: The natural logarithm $\ln(a)$ is the inverse of the exponential function, and it often appears in derivatives of exponential-like functions.