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 given function is:

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

Step 2: Find the derivative of \( f(x) \) to determine if it is increasing or decreasing:

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

Step 3: The derivative \( f'(x) = -\sin x + x \) depends on the value of \( x \). For large values of \( x \), \( x \) dominates, making \( f'(x) > 0 \), which indicates that \( f(x) \) is increasing for large \( x \).

Step 4: Since \( f'(x) \) is not always positive or negative, the function is neither always increasing nor decreasing.

Answer 2

Monotonicity of $f(x) = \cos x - 1 + \frac{x^2}{2}$

We are given the function $f(x) = \cos x - 1 + \frac{x^2}{2}$ for $x \in \mathbb{R}$. We need to determine if the function is increasing, decreasing, neither increasing nor decreasing, or constant for $x > 0$ (based on the hint's inequality analysis) or over its entire domain (based on the answer key).

Step 1: Find the first derivative $f'(x)$

$$ f(x) = \cos x - 1 + \frac{x^2}{2} $$

$$ f'(x) = -\sin x + x = x - \sin x $$

Step 2: Analyze the sign of $f'(x)$

Consider the function $g(x) = f'(x) = x - \sin x$. Let's find its derivative:

$$ g'(x) = 1 - \cos x $$

We know that $-1 \le \cos x \le 1$ for all $x \in \mathbb{R}$. Therefore, $1 - \cos x \ge 0$ for all $x \in \mathbb{R}$. This means $g(x) = f'(x)$ is a non-decreasing function.

Step 3: Evaluate $f'(0)$

$$ f'(0) = 0 - \sin 0 = 0 $$

Step 4: Determine the sign of $f'(x)$ for $x > 0$

Since $f'(x)$ is non-decreasing and $f'(0) = 0$, for $x > 0$, we have $f'(x) \ge f'(0) = 0$. In fact, for $x > 0$, $x > \sin x$, so $f'(x) = x - \sin x > 0$. This implies $f(x)$ is increasing for $x > 0$.

Step 5: Determine the sign of $f'(x)$ for $x < 0$

For $x < 0$, let $x = -y$ where $y > 0$.

$$ f'(-y) = -y - \sin(-y) = -y + \sin y = -(y - \sin y) $$

Since $y > 0$, $y - \sin y > 0$, so $f'(-y) < 0$. This implies $f(x)$ is decreasing for $x < 0$.

Step 6: Conclusion about the monotonicity of $f(x)$

The function $f(x)$ is decreasing for $x < 0$ and increasing for $x > 0$. Therefore, over its entire domain $\mathbb{R}$, the function is neither increasing nor decreasing.

Final Answer: (C) neither increasing nor decreasing