Suppose you want to use a Linear Approximation to approximate
1
25.2
without a calculator.

Which of the following functions and
𝑥
-values would be best to use?

Question

Suppose you want to use a Linear Approximation to approximate 
1
25.2
 without a calculator.

Which of the following functions and 
𝑥
-values would be best to use?

BlackTom AI · Accepted Answer

We start by restating the problem and the available options so we can compare them clearly.
Question: Suppose you want to use a Linear Approximation to approximate 1 25.2 without a calculator. Which of the following functions and x-values would be best to use?
Answer options:
1) Use f(x) = x and x = 9.
2) Use f(x) = x^(1/2) and x = 25.
3) Use f(x) = x and x = 25.1.
4) Use f(x) = x^(1/2) and x = 25.1.
5) Use f(x) = x and x = 25.
6) Use f(x) = 1/x and x = 9.

Option 1: Use f(x) = x and x = 9.
- This would be appropriate if you were approximating a value near x = 9 and using the linearization of f(x) = x, which is simply the identity line. However, the target value 25.2 is far from 9, and using a linear approximation of f(x) = x would not provide a meaningful improvement for sqrt-like targets. Additionally, for sqrt-type targets, the function is not x, so this choice misaligns the underlying function.

Option 2: Use f(x) = x^(1/2) and x = 25.
- This matches the standard technique: if you want to approximate a value close to 25 for the square root, you pick f(x) = sqrt(x) and a = 25. The derivative f'(x) = 1/(2 sqrt(x)) evaluated at a = 25 is 1/10, so the linearization L(x) = sqrt(25) + (x-25)/10. This gives a good approximation for x near 25, such as 25.1 or 25.2. This option aligns the function with the target type (square root) and uses a nearby a for a small, accurate correction.

Option 3: Use f(x) = x and x = 25.1.
- Again, choosing the identity function f(x) = x would be inappropriate for approximating a square root value. Although x = 25.1 is near the target value, the chosen function does not reflect the square-root relationship, so the linear approximation would not be the appropriate form, and accuracy would suffer for this context.

Option 4: Use f(x) = x^(1/2) and x = 25.1.
- This uses the correct function f(x) = sqrt(x) but centers at a = 25.1. While this is still reasonable, using a = 25 is typically preferred when the target value is around 25 (e.g., 25.2) because 25 is a neater, exact square root (sqrt(25) = 5) and yields a simple, clean derivative value of 1/10. Using a = 25.1 would complicate the arithmetic slightly without a substantial gain in accuracy for values near 25.2.

Option 5: Use f(x) = x and x = 25.
- This is the same as Option 3 but with a = 25. The problem is that f(x) = x does not model the square-root behavior; even near 25, using the identity function to approximate a square-root value would be inaccurate. Hence this choice is not suitable for a square-root target.

Option 6: Use f(x) = 1/x and x = 9.
- This uses the reciprocal function, which is unrelated to the square root and would not provide a meaningful linear approximation for a value like sqrt(25.2) or similar. The 1/x model would misrepresent the target’s local behavior, leading to large errors in this context.

In summary, the central idea is to choose a function that matches the type of quantity you’re approximating (a square root in this case) and to pick a base point a that is close to the target value so the linear approximation works well. Among the given options, Option 2 (f(x) = x^(1/2) and x = 25) best satisfies these criteria, because it uses the square-root function and a nearby expansion point with a simple derivative. The rest either use the wrong function for a square-root target or select a base point that’s less optimal for accuracy near 25.

Suppose you want to use a Linear Approximation to approximate 1 25.2 without a calculator. Which of the following functions and 𝑥 -values would be best to use?单项选择题

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