Question

The approximate value of \(\sqrt[3]{63}\) is

• A. \(3.9791\)
• B. \(3.8791\)
• C. \(3.9581\)
• D. \(4.0205\)

Hint:

Use differentials near 64: cube root of 64 is 4, derivative is 1/(3 x^(2/3)) = 1/48.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need an approximate value of \(\sqrt[3]{63}\). The nearest perfect cube is \(64 = 4^3\), so we use differentials.

Step 2: Key Formula or Approach:
For a small change \(\Delta x\), \(f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x\). Take \(f(x) = x^{1/3}\), so \(f'(x) = \frac{1}{3x^{2/3}}\).

Step 3: Choose values:
Let \(x = 64\) and \(\Delta x = -1\). Then \(f(64) = 4\) and \(f'(64) = \frac{1}{3 \times 16} = \frac{1}{48}\).

Step 4: Calculate:
\[ \sqrt[3]{63} \approx 4 + \frac{1}{48}(-1) = 4 - 0.0208 = 3.9792 \] A slightly more careful value (adding the second-order term) is 3.97906, which rounds to \(3.9791\).

Step 5: Check the options:
Options (B) 3.8791 and (C) 3.9581 are too far below 4. Option (D) 4.0205 is above 4, but \(\sqrt[3]{63}\) must be less than \(\sqrt[3]{64} = 4\). So (A) is the closest.

Final Answer:
The approximate value is 3.9791, option (A). \[ \boxed{3.9791} \]