Question

\(f(x) = \cos x - 1 + \frac{x^2}{2!}, \, x \in \mathbb{R}\)
Then \(f(x)\) is:

• A. decreasing function
• B. increasing function
• C. neither increasing nor decreasing
• D. constant for \(x>0\)

Hint:

When analyzing the behavior of a function, examine the sign of its derivative. If the derivative is not always positive or negative, the function may neither increase nor decrease uniformly.

Answer 1

The Correct Option is C

Step 1: The initial function is:

\[ f(x) = \cos x - 1 + \frac{x^2}{2!} \]

Step 2: Calculate the derivative of \( f(x) \) to analyze its behavior:

\[ f'(x) = -\sin x + x \]

Step 3: The derivative \( f'(x) = -\sin x + x \) depends on \( x \). For large \( x \), \( x \) primarily determines the sign, leading to \( f'(x) > 0 \), implying \( f(x) \) increases for large \( x \).

Step 4: Because \( f'(x) \) isn't consistently positive or negative, the function is neither always increasing nor always decreasing.