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To approximate \( \sqrt[3]{26} \) using differentials, we will first rewrite this as a function and apply the differential approximation method.

1. **Define the function**:
   Let \( f(x) = x^{1/3} \).

2. **Choose a nearby value for x**:
   Since we are approximating \( \sqrt[3]{26} \), we choose \( x = 25 \) because the cube root of 25 is close to that of 26 and simplifies calculations.

3. **Calculate the function and derivative at x = 25**:
   - \( f(25) = 25^{1/3} = 2.9240 \).
   - The derivative \( f'(x) = \frac{1}{3} x^{-2/3} \).
   - At \( x = 25 \), \( f'(25) = \frac{1}{3} \times 25^{-2/3} \approx 0.0390 \).

4. **Calculate the differential approximation**:
   We want to approximate \( f(26) \). Using differentials, we get:
   \[
   f(26) \approx f(25) + f'(25) \cdot (26 - 25)
   \]
   Substituting the values:
   \[
   f(26) \approx 2.9240 + 0.0390 \times 1 = 2.9630
   \]

So the corrected answer using differentials is approximately **2.9630**.

Would you like further details or explanations on any steps? Here are some related questions:

1. How do we choose the "nearby value" in differential approximations?
2. Why is the differential method useful for approximate calculations?
3. Can we use differentials to approximate values for more complex functions?
4. What are other applications of differentials in calculus?
5. How accurate is the differential method compared to other approximation methods?

**Tip**: Differential approximation works best when the point of expansion is very close to the target point, as errors increase with larger distances.

Use differentials to approximate the value of the expression. Compare your answer with that of a calculator. (Round your answers to four decimal places.)

Learn how to approximate the cube root of 26 using differentials. This solution explores key calculus concepts such as differential approximations and the derivative of cube root functions, suitable for students in Grades 11-12 or early college level.

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