Question:

The approximate value of $ f(5.001) $, $\text{where} $ $f(x) = x^3 - 7x^2 + 10 $

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To approximate function values for small changes in \( x \), use the linear approximation: \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \).
Updated On: May 2, 2025
  • -39.995
  • -38.995
  • -37.335
  • -40.995
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The Correct Option is A

Approach Solution - 1

We are given the function \( f(x) = x^3 - 7x^2 + 10 \) and need to approximate \( f(5.001) \). 
The first step is to approximate the value of the function using a linear approximation. We use the fact that the linear approximation to a function around a point \( x_0 \) is given by:
\[ f(x) \approx f(x_0) + f'(x_0)(x - x_0) \] 
Step 1: Calculate \( f(5) \)
\[ f(5) = 5^3 - 7(5^2) + 10 = 125 - 7 \cdot 25 + 10 = 125 - 175 + 10 = -40 \] 
Step 2: Find the derivative \( f'(x) \)
\[ f'(x) = 3x^2 - 14x \] 
Step 3: Calculate \( f'(5) \)
\[ f'(5) = 3(5^2) - 14(5) = 3 \cdot 25 - 70 = 75 - 70 = 5 \] 
Step 4: Apply the linear approximation
Now, use the approximation formula to estimate \( f(5.001) \): \[ f(5.001) \approx f(5) + f'(5)(5.001 - 5) = -40 + 5 \cdot 0.001 = -40 + 0.005 = -39.995 \] Thus, the approximate value of \( f(5.001) \) is \( -39.995 \).

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Approach Solution -2

To approximate the value of \( f(5.001) \) where \( f(x) = x^3 - 7x^2 + 10 \), we'll use linear approximation.

1. Set Up the Approximation:
We break \( 5.001 \) into \( x = 5 \) and \( \Delta x = 0.001 \), and use the formula:

\[ f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x) \]

2. Compute the Derivative:
First find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^3 - 7x^2 + 10) = 3x^2 - 14x \]

3. Apply the Approximation Formula:
Substitute into the approximation formula:

\[ f(5.001) \approx (5^3 - 7 \cdot 5^2 + 10) + 0.001 \cdot (3 \cdot 5^2 - 14 \cdot 5) \]

4. Calculate Each Component:
Compute each term step by step:
- \( 5^3 = 125 \)
- \( 7 \cdot 25 = 175 \)
- \( 3 \cdot 25 = 75 \)
- \( 14 \cdot 5 = 70 \)

5. Substitute and Simplify:
\[ f(5.001) \approx (125 - 175 + 10) + 0.001 \cdot (75 - 70) \]
\[ = (-40) + 0.001 \cdot 5 \]
\[ = -40 + 0.005 = -39.995 \]

Final Answer:
The approximate value is \( \boxed{-39.995} \).

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