To find the linear approximation of $(0.99)^{10}$, we use the concept of linear approximation based on a function's tangent line at a nearby point where the calculation is simpler.

Step-by-Step Solution:

1. Define the function: Let $f(x) = x^{10}$.

2. Choose a point near $x = 0.99$: We select $x = 1$ because it's close to $0.99$ and simplifies calculations.

3. Compute $f(x)$ and its derivative at $x = 1$:

   • $f(x) = x^{10}$
   • $f'(x) = 10x^9$

   At $x = 1$:

   • $f(1) = 1^{10} = 1$
   • $f'(1) = 10 \cdot 1^9 = 10$

4. Write the linear approximation formula: The linear approximation is: $f(x) \approx f(a) + f'(a)(x - a)$ where $a = 1$.

   Substituting the values: $f(x) \approx 1 + 10(x - 1)$

5. Approximate $f(0.99)$: Substitute $x = 0.99$: $f(0.99) \approx 1 + 10(0.99 - 1)$ $f(0.99) \approx 1 + 10(-0.01)$ $f(0.99) \approx 1 - 0.1 = 0.9$

Final Answer:

$(0.99)^{10} \approx 0.9$

Would you like more details or assistance with this topic?

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5 Related Questions:

1. How can we improve the accuracy of this approximation using higher-order terms?
2. What are some practical applications of linear approximations in real-world problems?
3. How does the error in linear approximation vary with the distance from the point of tangency?
4. Can you compute the exact value of $(0.99)^{10}$ and compare it to the approximation?
5. How would the approximation change if we used a point closer to $0.99$, like $0.995$?

Tip:

Linear approximation is most accurate when the chosen point is close to the value you're approximating, so selecting the right point is crucial!