Question

At \(x = 0\), \(f(x) = \cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3}\)

• A. has a minimum value
• B. has a maximum value
• C. has no extremum value
• D. is not defined

Hint:

If first derivative vanishes but second derivative is zero, check higher order derivatives; odd order non-zero indicates inflection point, not extremum.

Answer 1

The Correct Option is C

Step 1: Expand cos x using its Maclaurin series.
\[ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \]

Step 2: Substitute into f(x).
\[ f(x)=\cos x - 1 + \frac{x^2}{2} - \frac{x^3}{3} = \left(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\right)-1+\frac{x^2}{2}-\frac{x^3}{3} \]
The \(x^2\) terms cancel, leaving \[ f(x) = -\frac{x^3}{3} + \frac{x^4}{24} - \cdots \]

Step 3: Identify the dominant term near x = 0.
For small \(x\), the cubic term dominates over the quartic term, so \(f(x) \approx -\frac{x^3}{3}\) close to \(x=0\).

Step 4: Examine the sign change.
Since \(-\frac{x^3}{3}\) is positive for \(x \lt 0\) and negative for \(x \gt 0\), \(f(x)-f(0)\) changes sign as \(x\) passes through 0, so \(x=0\) cannot be a maximum or a minimum.

Step 5: Final conclusion.
Hence \(x=0\) gives \[ \boxed{\text{no extremum value}} \]