f(x)=cos (e^x - 2)
Linear Interpolation Method

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.

Learn how to apply the linear interpolation method to estimate the value of the function f(x) = cos(e^x - 2) between two known points. Includes step-by-step calculation and example for understanding. Suitable for advanced high school students and beyond.

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