## Step 1: Find the 4th Taylor Polynomial of \( f(x) = e^{3x} \) at \( x = \) The \( n \)th Taylor polynomial for \( f(x) \) centered at \( x = \) is: \[ T_n(x) = f() + f'()x + \frac{f''()}{2!}x^2 + \cdots + \frac{f^{(n)}()}{n!}x^n \] For \( f(x) = e^{3x} \): - \( f() = 1 \) - \( f'(x) = 3e^{3x} \implies f'() = 3 \) - \( f''(x) = 9e^{3x} \implies f''() = 9 \) - \( f'''(x) = 27e^{3x} \implies f'''() = 27 \) - \( f^{(4)}(x) = 81e^{3x} \implies f^{(4)}() = 81 \) So, \[ T_4(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 + \frac{81}{24}x^4 \] **Explanation:** We have constructed the 4th Taylor polynomial by calculating derivatives and evaluating them at \( x = \). --- ## Step 2: Use the Alternating Series Error Bound The error for the Taylor approximation is bounded by the next term: \[ |R_4(x)| \leq \left| \frac{f^{(5)}()}{5!}x^5 \right| \] Since \( f^{(5)}(x) = 243e^{3x} \), at \( x = \): \( f^{(5)}() = 243 \): \[ |R_4(x)| \leq \left| \frac{243}{120} x^5 \right| \] Set this less than or equal to \( .0001 \): \[ \frac{243}{120} x^5 \leq .0001 \] \[ x^5 \leq \frac{.0001 \times 120}{243} \] \[ x^5 \leq \frac{.012}{243} \] \[ x^5 \leq .0000493827 \] \[ x \leq (.0000493827)^{1/5} \] \[ x \leq .145 \] **Explanation:** We used the remainder bound for the Taylor polynomial, set it less than \( .0001 \), and solved for \( x \). --- ## **Final Answer** \[ \boxed{ < x \leq .145} \]

# Step 1: Find the 4th Taylor Polynomial of \( f(x) = e^{3x} \) at \( x = 0 \) The \( n \)th Taylor polynomial for \( f(x) \) centered at \( x = 0 \) is given by: \[ T_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 \] Calculating the necessary derivatives for \( f(x) = e^{3x} \): - \( f(0) = e^{0} = 1 \) - \( f'(x) = 3e^{3x} \implies f'(0) = 3 \) - \( f''(x) = 9e^{3x} \implies f''(0) = 9 \) - \( f'''(x) = 27e^{3x} \implies f'''(0) = 27 \) - \( f^{(4)}(x) = 81e^{3x} \implies f^{(4)}(0) = 81 \) Thus, the 4th Taylor polynomial is: \[ T_4(x) = 1 + 3x + \frac{9}{2}x^2 + \frac{27}{6}x^3 + \frac{81}{24}x^4 \] **Explanation:** We derived the 4th Taylor polynomial by evaluating the function and its derivatives at \( x = 0 \). --- # Step 2: Use the Alternating Series Error Bound The error for the Taylor approximation is given by the next term: \[ |R_4(x)| \leq \left| \frac{f^{(5)}(0)}{5!}x^5 \right| \] Calculating \( f^{(5)}(x) \): - \( f^{(5)}(x) = 243e^{3x} \implies f^{(5)}(0) = 243 \) Thus, the error bound becomes: \[ |R_4(x)| \leq \left| \frac{243}{120} x^5 \right| \] We want this to be less than or equal to \( 0.0001 \): \[ \frac{243}{120} x^5 \leq 0.0001 \] Rearranging gives: \[ x^5 \leq \frac{0.0001 \times 120}{243} \] \[ x^5 \leq \frac{0.012}{243} \] \[ x^5 \leq 0.0000493827 \] Taking the fifth root: \[ x \leq (0.0000493827)^{1/5} \] Calculating gives: \[ x \leq 0.145 \] **Explanation:** We applied the remainder bound for the Taylor polynomial, set the upper limit, and solved for \( x \). --- ## **Final Answer** \[ \boxed{0 < x \leq 0.145} \]