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The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 is
- a)3x2 − 6x − 5
- b)−3x2 − 5
- c)−3x2 + 6x − 5
- d)3x2 – 5
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 i...
Concept:
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
It is also called Maclaurin's series.
Calculation:
Calculation:
f(x) = x3 - 3x2-5
f(0) = 03 - 3 × 02 - 5 = - 5
f'(0) = 3x2 - 6x = 0
f"(0) = 6x - 6 = - 6
The quadratic approximation of f(x) at the point x = 0 is
Most Upvoted Answer
The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 i...
Understanding Quadratic Approximation
The quadratic approximation of a function at a point gives a polynomial that closely resembles the function near that point. For a function \( f(x) \), the quadratic approximation at \( x = a \) is given by:
\[ f(x) \approx f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2 \]
Step 1: Calculate \( f(x) \)
For the function
\[ f(x) = x^3 - 3x^2 - 5 \]
we will find \( f(0) \), \( f'(0) \), and \( f''(0) \).
Step 2: Find \( f(0) \)
- Calculate \( f(0) \):
\[ f(0) = 0^3 - 3(0^2) - 5 = -5 \]
Step 3: Find the first derivative \( f'(x) \)
- The first derivative is:
\[ f'(x) = 3x^2 - 6x \]
- Now, evaluate at \( x = 0 \):
\[ f'(0) = 3(0^2) - 6(0) = 0 \]
Step 4: Find the second derivative \( f''(x) \)
- The second derivative is:
\[ f''(x) = 6x - 6 \]
- Now, evaluate at \( x = 0 \):
\[ f''(0) = 6(0) - 6 = -6 \]
Step 5: Construct the Quadratic Approximation
Now, substituting back into the quadratic approximation formula around \( x = 0 \):
- Quadratic approximation:
\[ f(x) \approx -5 + 0 \cdot x + \frac{-6}{2}(x^2) \]
- This simplifies to:
\[ f(x) \approx -5 - 3x^2 \]
Thus, the quadratic approximation at \( x = 0 \) is:
-3x^2 - 5
Hence, the correct answer is option B: \(-3x^2 - 5\).
The quadratic approximation of a function at a point gives a polynomial that closely resembles the function near that point. For a function \( f(x) \), the quadratic approximation at \( x = a \) is given by:
\[ f(x) \approx f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2 \]
Step 1: Calculate \( f(x) \)
For the function
\[ f(x) = x^3 - 3x^2 - 5 \]
we will find \( f(0) \), \( f'(0) \), and \( f''(0) \).
Step 2: Find \( f(0) \)
- Calculate \( f(0) \):
\[ f(0) = 0^3 - 3(0^2) - 5 = -5 \]
Step 3: Find the first derivative \( f'(x) \)
- The first derivative is:
\[ f'(x) = 3x^2 - 6x \]
- Now, evaluate at \( x = 0 \):
\[ f'(0) = 3(0^2) - 6(0) = 0 \]
Step 4: Find the second derivative \( f''(x) \)
- The second derivative is:
\[ f''(x) = 6x - 6 \]
- Now, evaluate at \( x = 0 \):
\[ f''(0) = 6(0) - 6 = -6 \]
Step 5: Construct the Quadratic Approximation
Now, substituting back into the quadratic approximation formula around \( x = 0 \):
- Quadratic approximation:
\[ f(x) \approx -5 + 0 \cdot x + \frac{-6}{2}(x^2) \]
- This simplifies to:
\[ f(x) \approx -5 - 3x^2 \]
Thus, the quadratic approximation at \( x = 0 \) is:
-3x^2 - 5
Hence, the correct answer is option B: \(-3x^2 - 5\).
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Community Answer
The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 i...
Concept:
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
The Taylor's series expansion of f(x) about origin (i.e x = 0) is given by
f(x) = f(0) + x × f′(0) + x2/2! × f"(0)+.....
It is also called Maclaurin's series.
Calculation:
Calculation:
f(x) = x3 - 3x2-5
f(0) = 03 - 3 × 02 - 5 = - 5
f'(0) = 3x2 - 6x = 0
f"(0) = 6x - 6 = - 6
The quadratic approximation of f(x) at the point x = 0 is
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The quadratic approximation of (x) = x3 - 3x2 - 5 at the point x = 0 isa)3x2 − 6x − 5b)−3x2 − 5c)−3x2 + 6x − 5d)3x2 – 5Correct answer is option 'B'. Can you explain this answer?
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