1. **State the problem:** We are given two points on the function $n(x)$: $n(1.36) = 0.9131$ and $n(1.37) = 0.9147$. We want to approximate $n(1.367)$ and $n(1.356)$ using linear interpolation or extrapolation. 2. **Formula for linear interpolation/extrapolation:** $$n(x) \approx n(x_0) + \frac{n(x_1) - n(x_0)}{x_1 - x_0} (x - x_0)$$ where $x_0$ and $x_1$ are known points, and $x$ is the point where we want to estimate $n(x)$. 3. **Calculate the slope:** $$m = \frac{0.9147 - 0.9131}{1.37 - 1.36} = \frac{0.0016}{0.01} = 0.16$$ 4. **Approximate $n(1.367)$ (interpolation since $1.367$ is between $1.36$ and $1.37$):** $$n(1.367) \approx 0.9131 + 0.16 \times (1.367 - 1.36) = 0.9131 + 0.16 \times 0.007 = 0.9131 + 0.00112 = 0.91422$$ 5. **Approximate $n(1.356)$ (extrapolation since $1.356$ is less than $1.36$):** $$n(1.356) \approx 0.9131 + 0.16 \times (1.356 - 1.36) = 0.9131 + 0.16 \times (-0.004) = 0.9131 - 0.00064 = 0.91246$$ **Final answers:** - $n(1.367) \approx 0.91422$ - $n(1.356) \approx 0.91246$