Let's go through each part of the problem step-by-step:

Part (a): Write the equation of the linear approximation

Given:

• $f(x) = \cos(x)$
• Point $a = \frac{\pi}{2}$

1. Find $f(a)$: $f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

2. Find $f'(x)$: The derivative of $f(x) = \cos(x)$ is: $f'(x) = -\sin(x)$

3. Evaluate $f'(a)$: $f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$

4. Construct the linear approximation $L(x)$: The linear approximation (tangent line) at $x = a$ is: $L(x) = f(a) + f'(a)(x - a)$ Substituting the values we found: $L(x) = 0 + (-1)(x - \frac{\pi}{2}) = -x + \frac{\pi}{2}$ So, the equation of the linear approximation is: $L(x) = \frac{\pi}{2} - x$

Part (b): Use the linear approximation to estimate $f(1.53)$

Using our linear approximation $L(x) = \frac{\pi}{2} - x$, we substitute $x = 1.53$: $L(1.53) = \frac{\pi}{2} - 1.53$ Using $\frac{\pi}{2} \approx 1.5708$: $L(1.53) \approx 1.5708 - 1.53 = 0.0408$

Thus, the estimated value of $f(1.53)$ using the linear approximation is approximately $0.0408$.

Part (c): Compute the percent error

1. Find the exact value of $f(1.53)$: $f(1.53) = \cos(1.53) \approx 0.0423$

2. Calculate the percent error: The formula for percent error is: $\text{Percent Error} = 100 \times \frac{| \text{approximation} - \text{exact} |}{|\text{exact}|}$ Substituting the values: $\text{Percent Error} = 100 \times \frac{|0.0408 - 0.0423|}{|0.0423|}$ $= 100 \times \frac{0.0015}{0.0423} \approx 3.55\%$

Summary of Answers

1. Linear Approximation Equation: $L(x) = \frac{\pi}{2} - x$
2. Estimated Value at $x = 1.53$: $L(1.53) \approx 0.0408$
3. Percent Error: Approximately $3.55\%$

Would you like further clarification on any step?

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Expanding Questions

1. What is the process for finding the linear approximation of a function at any given point?
2. How does the choice of point $a$ affect the accuracy of a linear approximation?
3. Why do we use the derivative to find the slope in a linear approximation?
4. How would the approximation change if we were to use a different function, such as $\sin(x)$, at the same point?
5. How does the error percentage reflect the quality of an approximation?

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Tip: Linear approximations work best when $x$ is close to the point $a$ because the tangent line more accurately represents the function’s behavior near that point.