Consider the following.

Question

f(x) = x4 − x + 9
Compare the values of Δy and dy if x changes from 1 to 1.04. (Round your answers to four decimal places.)
Δy =
dy =
What if x changes from 1 to 1.02? (Round your answers to four decimal places.)
Δy =
dy =
Does the approximation
Δy ≈ dy
become better as Δx gets smaller?

Yes
No

Answer 1

To analyze the changes in the function \( f(x) = x^4 - x + 9 \) as \( x \) changes, we will calculate both \( \Delta y \) and \( dy \) for the specified intervals.

Step 1: Calculate \( \Delta y \)

1. Find \( f(1) \): \[ f(1) = 1^4 - 1 + 9 = 1 - 1 + 9 = 9 \]

2. Find \( f(1.04) \): \[ f(1.04) = (1.04)^4 - (1.04) + 9 \] Calculating \( (1.04)^4 \): \[ (1.04)^4 = 1.16985856 \] Then: \[ f(1.04) = 1.16985856 - 1.04 + 9 = 1.16985856 - 1.04 + 9 \approx 9.12985856 \]

3. Calculate \( \Delta y \): \[ \Delta y = f(1.04) - f(1) = 9.12985856 - 9 = 0.12985856 \approx 0.1299 \]

Step 2: Calculate \( dy \)

1. Calculate \( \frac{df}{dx} \): \[ f'(x) = 4x^3 - 1 \] Evaluating at \( x = 1 \): \[ f'(1) = 4(1^3) - 1 = 4 - 1 = 3 \]

2. Calculate \( dy \), where \( \Delta x = 1.04 - 1 = 0.04 \): \[ dy = f'(1) \cdot \Delta x = 3 \cdot 0.04 = 0.12 \]

Summary for \( x = 1 \) to \( x = 1.04 \):

\[ \Delta y \approx 0.1299, \quad dy = 0.12 \]

Step 3: Now, calculate for \( x \) changes from 1 to 1.02

1. Find \( f(1.02) \): \[ f(1.02) = (1.02)^4 - (1.02) + 9 \] Calculating \( (1.02)^4 \): \[ (1.02)^4 \approx 1.08285632 \] Then: \[ f(1.02) \approx 1.08285632 - 1.02 + 9 \approx 9.06285632 \]

2. Calculate \( \Delta y \): \[ \Delta y = f(1.02) - f(1) = 9.06285632 - 9 = 0.06285632 \approx 0.0629 \]

3. Calculate \( dy \), where \( \Delta x = 1.02 - 1 = 0.02 \): \[ dy = f'(1) \cdot \Delta x = 3 \cdot 0.02 = 0.06 \]

Summary for \( x = 1 \) to \( x = 1.02 \):

\[ \Delta y \approx 0.0629, \quad dy = 0.06 \]

Final Results:

1. For \( x = 1 \) to \( x = 1.04 \):

   • \( \Delta y \approx 0.1299 \)
   • \( dy = 0.12 \)

2. For \( x = 1 \) to \( x = 1.02 \):

   • \( \Delta y \approx 0.0629 \)
   • \( dy = 0.06 \)

Conclusion:

The approximation \( \Delta y \approx dy \) does indeed become better as \( \Delta x \) gets smaller.

Answer:

• \( \Delta y \) from 1 to 1.04: \( 0.1299 \)
• \( dy \) from 1 to 1.04: \( 0.12 \)
• \( \Delta y \) from 1 to 1.02: \( 0.0629 \)
• \( dy \) from 1 to 1.02: \( 0.06 \)

Does the approximation \( \Delta y \approx dy \) become better as \( \Delta x \) gets smaller?

• Yes.