Questions: Analyze the concavity and find the inflection points of f(x) = cos(x^2).

Solution Steps

To analyze the concavity and find the inflection points of the function \( f(x) = \cos(x^2) \), we need to find the second derivative of the function. Inflection points occur where the second derivative changes sign, which indicates a change in concavity. We will:

1. Compute the first derivative \( f'(x) \).
2. Compute the second derivative \( f''(x) \).
3. Solve \( f''(x) = 0 \) to find potential inflection points.
4. Check the sign change of \( f''(x) \) around these points to confirm inflection points.

To analyze the concavity and find the inflection points of the function \( f(x) = \cos(x^2) \), we first need to find the first derivative \( f'(x) \).

Using the chain rule, the derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \), and the derivative of \( x^2 \) with respect to \( x \) is \( 2x \). Therefore, the first derivative is:

\[ f'(x) = \frac{d}{dx} \left[ \cos(x^2) \right] = -\sin(x^2) \cdot 2x = -2x \sin(x^2) \]

Next, we find the second derivative \( f''(x) \) to analyze the concavity.

Using the product rule and chain rule, the derivative of \( -2x \sin(x^2) \) is:

\[ f''(x) = \frac{d}{dx} \left[ -2x \sin(x^2) \right] \]

Applying the product rule:

\[ f''(x) = -2 \sin(x^2) + (-2x) \cdot \frac{d}{dx}[\sin(x^2)] \]

The derivative of \( \sin(x^2) \) is \( \cos(x^2) \cdot 2x \), so:

\[ f''(x) = -2 \sin(x^2) - 4x^2 \cos(x^2) \]

Inflection points occur where the second derivative changes sign, which means we need to solve \( f''(x) = 0 \):

\[ -2 \sin(x^2) - 4x^2 \cos(x^2) = 0 \]

Factor out the common term:

\[ -2 (\sin(x^2) + 2x^2 \cos(x^2)) = 0 \]

This implies:

\[ \sin(x^2) + 2x^2 \cos(x^2) = 0 \]

This equation is complex to solve analytically, but we can analyze it for potential solutions. The inflection points are where the expression changes sign, which typically requires numerical methods or graphing to find specific values of \( x \).

Final Answer