Question

The approximate value of \( f(5.001) \), where \( f(x) = x^3 - 7x^2 + 15 \), is:

• A. -34.995
• B. -33.995
• C. -33.335
• D. -35.995

Hint:

For small changes in \( x \), use the first-order approximation \( f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x \).

Answer 1

The Correct Option is A

We are given the function \( f(x) = x^3 - 7x^2 + 15 \). We need to find the approximate value of \( f(5.001) \). Step 1: Use the first-order approximation The first-order approximation for \( f(x) \) near \( x = 5 \) is: \[ f(5.001) \approx f(5) + f'(5) \cdot 0.001 \] Step 2: Calculate \( f(5) \) and \( f'(x) \) First, find \( f(5) \): \[ f(5) = 5^3 - 7 \cdot 5^2 + 15 = 125 - 175 + 15 = -35 \] Next, find the derivative \( f'(x) \): \[ f'(x) = 3x^2 - 14x \] At \( x = 5 \): \[ f'(5) = 3 \cdot 5^2 - 14 \cdot 5 = 75 - 70 = 5 \] Step 3: Approximate \( f(5.001) \) Now, using the approximation: \[ f(5.001) \approx f(5) + f'(5) \cdot 0.001 = -35 + 5 \cdot 0.001 = -34.995 \] Thus, the approximate value of \( f(5.001) \) is \( -34.995 \).