To estimate the maximum error in the approximation $e^x \approx 1 + x + \frac{x^2}{2}$ over the interval $\left[-\frac{3}{7}, \frac{3}{7}\right]$, we use the remainder term from the Taylor series expansion of $e^x$.

The Taylor series expansion of $e^x$ around $x = 0$ is: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$

The remainder term $R_2(x)$ after the second term is given by: $R_2(x) = \frac{e^c \cdot x^3}{6}$ where $c$ is some number between $0$ and $x$.

To find the maximum error, we need to consider the maximum value of $|R_2(x)|$ on the interval $\left[-\frac{3}{7}, \frac{3}{7}\right]$.

Since $e^c$ is an increasing function and $|x|$ is maximized at $\frac{3}{7}$ in the given interval, we get: $|R_2(x)| \leq \frac{e^{\frac{3}{7}} \cdot \left(\frac{3}{7}\right)^3}{6}$

First, calculate $e^{\frac{3}{7}}$: $e^{\frac{3}{7}} \approx 1.428$

Now calculate the remainder term: $R_2\left(\frac{3}{7}\right) = \frac{1.428 \cdot \left(\frac{3}{7}\right)^3}{6}$ $\left(\frac{3}{7}\right)^3 = \frac{27}{343}$ $R_2\left(\frac{3}{7}\right) = \frac{1.428 \cdot \frac{27}{343}}{6}$ $R_2\left(\frac{3}{7}\right) = \frac{1.428 \cdot 27}{2058}$ $R_2\left(\frac{3}{7}\right) \approx \frac{38.556}{2058}$ $R_2\left(\frac{3}{7}\right) \approx 0.0187$

Therefore, the maximum error is approximately $0.0187$.

Since we need to use scientific notation and round to two decimal places: $0.0187 \approx 1.87 \times 10^{-2}$

So, the maximum error is approximately $1.87 \times 10^{-2}$.

Would you like more details or have any questions?

Here are five questions you might consider asking next:

1. Can you explain the Taylor series expansion in more detail?
2. How do you determine the remainder term in a Taylor series?
3. What is the significance of the interval in the error estimation?
4. How does the value of $c$ affect the remainder term?
5. Can you show a step-by-step calculation for another interval?

Tip: When approximating functions using Taylor series, always consider higher-order terms to understand the accuracy of your approximation better.