Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is473754\frac{73}{75}.52125\frac{2}{12}.42325.4\frac{23}{25}.497100.4\frac{97}{100}.41315.4\frac{13}{15}.

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Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is473754\frac{73}{75}.52125\frac{2}{12}.42325.4\frac{23}{25}.497100.4\frac{97}{100}.41315.4\frac{13}{15}.

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Question restatement: Using differentials, approximate the cube root of 123, i.e., ∛123, with the given options for the result.

Option 1: 4 73/75. This matches the differential approximation: take a nearby perfect cube, here 125 = 5^3, so ∛125 = 5. Let f(x) = x^{1/3}; f'(x) = (1/3)x^{-2/3}. At x0 = 125, f'(125) = (1/3) * 125^{-2/3} = (1/3) * (1/25) = 1/75. Since 123 = 125 - 2, the differential gives ∛123 ≈ ∛125 + f'(125)(-2) = 5 - 2/75 = (375 - 2)/75 = 373/75 = 4 73/75. This exactly corresponds to the first option and aligns with the standard differential approach.

Option 2: 5 2/12. This equals 5 + 2/12 ≈ 5.166..., which is far from the actual 4.97 range and ignores the correct sign and magnitude of the adjustment from 125 to 123. It does not follow from the tangent line approximation at x0 = 125.

Option 3: 4 23/25. This equals 4.92. While it is in the ballpark (near 4.97), it does not reflect the precise linear correction computed with f'(125) = 1/75 and Δx = -2. The fractional adjustment 23/25 would imply a larger downward shift than the correct -2/75 term yields when added to 5.

Option 4: 4 97/100. This equals 4.97 exactly, but as a mixed number it would be 4 + 97/100 = 4.97, which is a rounded form that coincidentally resembles the target. However, the standard differential calculation gives 373/75, not 4.97 as a decimal with that level of rounding, so this option does not reflect the exact tangent-line-based approximation from the method.

Option 5: 4 13/15. This equals 4.866..., which is an undershoot compared to the tangent-line approximation. It represents a larger downward adjustment than the actual derivative-based change would produce when moving from 125 to 123.

Summary of reasoning across options: The correct differential approach uses the nearby cube 125 with ∛125 = 5 and f'(125) = 1/75, then Δx = -2 because 123 = 125 - 2. The predicted change is f'(125)Δx = (-2)/75, giving ∛123 ≈ 5 - 2/75 = 373/75 = 4 73/75. The other options either use incorrect Δx, incorrect derivative value, or incorrect sign, leading to numbers like 5.166..., 4.92, 4.97 with a rounding that does not originate from the precise linear approximation, or 4.866..., none of which match the standard differential result.

Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is473754\frac{73}{75}.52125\frac{2}{12}.42325.4\frac{23}{25}.497100.4\frac{97}{100}.41315.4\frac{13}{15}.

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Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is497100.4\frac{97}{100}.41315.4\frac{13}{15}.42325.4\frac{23}{25}.52125\frac{2}{12}.473754\frac{73}{75}.

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Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is473754\frac{73}{75}.52125\frac{2}{12}.42325.4\frac{23}{25}.497100.4\frac{97}{100}.41315.4\frac{13}{15}.

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at 的线性近似值为： 提示：at 的线性近似值为： 。

The linear approximation of 𝑓 ( 𝑥 ) = 5 𝑥 − 4 2 𝑥 2 + 1 at 𝑥 = − 3 is: Hint: The linear approximation of 𝑦 = 𝑓 ( 𝑥 ) at 𝑥 = 𝑥 0 is: 𝐿 ( 𝑥 ) = 𝑓 ′ ( 𝑥 0 ) ( 𝑥 − 𝑥 0 ) + 𝑓 ( 𝑥 0 ) .

This is a continuation of the previous question. What is the Linear Approximation of 𝑓 ( 𝑥 ) = tan ⁡ 𝑥 at 𝑥 = 𝜋 4 ?

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Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is497100.4\frac{97}{100}.41315.4\frac{13}{15}.42325.4\frac{23}{25}.52125\frac{2}{12}.473754\frac{73}{75}.

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at 的线性近似值为：提示：at 的线性近似值为：。