1. **State the problem:** We are given the degree 4 Taylor polynomial $P_4(x) = 9 + \frac{1}{7}(x-2)^3 + 7(x-2)^4$ for a function $f(x)$ centered at $x=2$. We want to find an upper bound on the error $|P_4(1.5) - f(1.5)|$ using the Lagrange error bound. 2. **Recall the Lagrange error bound formula:** $$ |R_n(x)| = |f(x) - P_n(x)| \leq \frac{M}{(n+1)!} |x - a|^{n+1} $$ where: - $n=4$ (degree of polynomial) - $a=2$ (center of expansion) - $M$ is an upper bound on $|f^{(n+1)}(z)|$ for $z$ between $a$ and $x$ 3. **Identify the 5th derivative bound $M$:** From the graph of $f^{(5)}(x)$, the values between $x=1.5$ and $x=2$ range approximately from $-3$ to about $4$ (since the graph decreases steeply near 2 to about -3 and is above 4 near -0.8). The maximum absolute value in this interval is $M = 4$. 4. **Calculate the error bound:** - $n+1 = 5$ - $|x - a| = |1.5 - 2| = 0.5$ - $5! = 120$ So, $$ |R_4(1.5)| \leq \frac{4}{120} (0.5)^5 $$ 5. **Simplify the expression:** $$ (0.5)^5 = 0.5 \times 0.5 \times 0.5 \times 0.5 \times 0.5 = 0.03125 $$ Therefore, $$ |R_4(1.5)| \leq \frac{4}{120} \times 0.03125 = \frac{4 \times 0.03125}{120} $$ 6. **Cancel common factors:** $$ \frac{4 \times 0.03125}{120} = \frac{\cancel{4} \times 0.03125}{\cancel{120} \times 30} = \frac{0.03125}{30} $$ 7. **Final calculation:** $$ \frac{0.03125}{30} = 0.0010416667 $$ **Answer:** The upper bound on the error $|P_4(1.5) - f(1.5)|$ is approximately $0.00104$.