Question:
At \(x = 0\), \(f(x) = \cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3}\)
At \(x = 0\), \(f(x) = \cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3}\)
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If first derivative vanishes but second derivative is zero, check higher order derivatives; odd order non-zero indicates inflection point, not extremum.
Updated On: Jul 18, 2026
- has a minimum value
- has a maximum value
- has no extremum value
- is not defined
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The Correct Option is C
Solution and Explanation
Step 1: Find first derivative.
\(f(x) = \cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3}\)
\[ f'(x) = -\sin x + x - x^2 \]
Step 2: Evaluate at \(x=0\).
\[ f'(0) = -\sin 0 + 0 - 0 = 0 \]
So \(x = 0\) is a critical point.
Step 3: Find second derivative.
\[ f''(x) = -\cos x + 1 - 2x \]
Step 4: Evaluate second derivative at \(x=0\).
\[ f''(0) = -1 + 1 - 0 = 0 \]
Second derivative test is inconclusive.
Step 5: Consider higher order terms.
Check third derivative: \(f'''(x) = \sin x - 2\)
\[ f'''(0) = 0 - 2 = -2 \neq 0 \]
Odd order derivative non-zero implies no extremum at \(x=0\).
Step 6: Final conclusion.
Hence, \(f(x)\) has no extremum at \(x=0\) \[ \boxed{\text{has no extremum value}} \]
\(f(x) = \cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3}\)
\[ f'(x) = -\sin x + x - x^2 \]
Step 2: Evaluate at \(x=0\).
\[ f'(0) = -\sin 0 + 0 - 0 = 0 \]
So \(x = 0\) is a critical point.
Step 3: Find second derivative.
\[ f''(x) = -\cos x + 1 - 2x \]
Step 4: Evaluate second derivative at \(x=0\).
\[ f''(0) = -1 + 1 - 0 = 0 \]
Second derivative test is inconclusive.
Step 5: Consider higher order terms.
Check third derivative: \(f'''(x) = \sin x - 2\)
\[ f'''(0) = 0 - 2 = -2 \neq 0 \]
Odd order derivative non-zero implies no extremum at \(x=0\).
Step 6: Final conclusion.
Hence, \(f(x)\) has no extremum at \(x=0\) \[ \boxed{\text{has no extremum value}} \]
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