Question:
The approximate value of $f(5.001)$, where $f(x)=x^3 - 7x^2 + 15$, is
The approximate value of $f(5.001)$, where $f(x)=x^3 - 7x^2 + 15$, is
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For small change, always use $f(x+h) \approx f(x)+hf'(x)$.
Updated On: Apr 23, 2026
- $-34.995$
- $-33.995$
- $-33.335$
- $-35.993$
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The Correct Option is A
Solution and Explanation
Concept:
Use linear approximation:
\[
f(x+h) \approx f(x) + hf'(x)
\]
Step 1: Take $x=5,\ h=0.001$.
Step 2: Compute $f(5)$.
\[ = 125 - 175 + 15 = -35 \]
Step 3: Find derivative.
\[ f'(x)=3x^2 -14x \] \[ f'(5)=75 - 70 = 5 \]
Step 4: Apply approximation.
\[ f(5.001) \approx -35 + 0.001 \times 5 = -35 + 0.005 = -34.995 \] Conclusion:
Answer = $-34.995$
Step 1: Take $x=5,\ h=0.001$.
Step 2: Compute $f(5)$.
\[ = 125 - 175 + 15 = -35 \]
Step 3: Find derivative.
\[ f'(x)=3x^2 -14x \] \[ f'(5)=75 - 70 = 5 \]
Step 4: Apply approximation.
\[ f(5.001) \approx -35 + 0.001 \times 5 = -35 + 0.005 = -34.995 \] Conclusion:
Answer = $-34.995$
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