Linear interpolation is a method to estimate the value of a function $f(x)$ at a point $x$ using known values of the function at two points, say $x_1$ and $x_2$. The formula for linear interpolation between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$f(x) \approx y_1 + \frac{y_2 - y_1}{x_2 - x_1} \cdot (x - x_1)$

For the function $f(x) = \cos(e^x - 2)$, let's assume you want to interpolate the value of $f(x)$ at some point $x$ between two known points $x_1$ and $x_2$.

Step-by-Step Linear Interpolation:

1. Calculate $y_1$ and $y_2$: $y_1 = f(x_1) = \cos(e^{x_1} - 2)$ $y_2 = f(x_2) = \cos(e^{x_2} - 2)$

2. Use the linear interpolation formula: $f(x) \approx y_1 + \frac{y_2 - y_1}{x_2 - x_1} \cdot (x - x_1)$

Example Calculation:

Let's assume $x_1 = 0$ and $x_2 = 1$, and you want to interpolate $f(x)$ at $x = 0.5$.

1. Calculate $y_1$ and $y_2$: $y_1 = \cos(e^0 - 2) = \cos(1 - 2) = \cos(-1) \approx 0.5403$ $y_2 = \cos(e^1 - 2) = \cos(e - 2) \approx \cos(0.7183) \approx 0.7539$

2. Linear interpolation at $x = 0.5$: $f(0.5) \approx 0.5403 + \frac{0.7539 - 0.5403}{1 - 0} \cdot (0.5 - 0)$ $f(0.5) \approx 0.5403 + 0.2136 \cdot 0.5 = 0.5403 + 0.1068 = 0.6471$

So, the interpolated value of $f(0.5)$ is approximately $0.6471$.

Would you like more details or have any questions?

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Here are five related questions:

1. What is the difference between linear and polynomial interpolation?
2. How does the accuracy of linear interpolation compare to higher-order methods?
3. Can you apply linear interpolation to non-continuous functions?
4. How would you perform linear interpolation if given three points instead of two?
5. What are some real-world applications of linear interpolation?

Tip: Linear interpolation is most accurate when the function is nearly linear over the interval between $x_1$ and $x_2$. For functions with significant curvature, consider higher-order interpolation methods.