A function is `y=x^(3)`. The approximate value of the function at `x=2.0001` is possibly (choose the closest answer possible)

Options

A. 8.0006
B. 8.0008
C. 8.0012
D. 8.0015

Answer

To find the approximate value of the function \( y = x^3 \) at \( x = 2.0001 \), we can use the concept of differentials. Here’s a step-by-step solution:

### Step 1: Identify the function and the point of interest
The function is given as:
\[ y = x^3 \]
We need to approximate the value of this function at \( x = 2.0001 \).

### Step 2: Express \( x \) in terms of a small increment
We can express \( x \) as:
\[ x = 2 + 0.0001 \]
where \( \Delta x = 0.0001 \).

### Step 3: Differentiate the function
Next, we differentiate the function \( y = x^3 \):
\[ \frac{dy}{dx} = 3x^2 \]

### Step 4: Calculate \( \Delta y \)
Using the differential approximation, we have:
\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]
Substituting \( x = 2 \) and \( \Delta x = 0.0001 \):
\[ \Delta y \approx 3(2^2)(0.0001) = 3(4)(0.0001) = 0.0012 \]

### Step 5: Calculate the value of \( y \) at \( x = 2 \)
Now, we calculate \( y \) at \( x = 2 \):
\[ y(2) = 2^3 = 8 \]

### Step 6: Find the approximate value of \( y \) at \( x = 2.0001 \)
Finally, we can find the approximate value of \( y \) at \( x = 2.0001 \):
\[ y(2.0001) \approx y(2) + \Delta y = 8 + 0.0012 = 8.0012 \]

### Conclusion
Thus, the approximate value of the function \( y = x^3 \) at \( x = 2.0001 \) is:
\[ \boxed{8.0012} \]