Question

Let

\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]

Then \( f'(-4) \) is equal to:

• A. \( \frac{\pi - 8}{2} \)
• B. \( \frac{16 + \pi}{2} \)
• C. \( \frac{8 + \pi}{2} \)
• D. \( \frac{\pi - 16}{2} \)
• E. \( \pi - 16 \)

Hint:

When differentiating piecewise functions, always identify which case applies to the given value of \( x \), then apply the appropriate rules (such as the product rule) to differentiate the function.

Answer 1

The Correct Option is B

We are given the piecewise function:

\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]

We are asked to find \( f'(-4) \).
Since \( -4 < 0 \), we will use the second case of the piecewise function:

\[ f(x) = x\left( \frac{\pi}{2} - x \right) \]

Step 1: Differentiate the function
Differentiate \( f(x) = x\left( \frac{\pi}{2} - x \right) \) using the product rule:

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

The product rule states that:

\[ f'(x) = \frac{d}{dx} (x) \cdot \left( \frac{\pi}{2} - x \right) + x \cdot \frac{d}{dx} \left( \frac{\pi}{2} - x \right) \]

Now calculate the derivatives:

\[ \frac{d}{dx} (x) = 1 \quad \text{and} \quad \frac{d}{dx} \left( \frac{\pi}{2} - x \right) = -1 \]

Thus:

\[ f'(x) = 1 \cdot \left( \frac{\pi}{2} - x \right) + x \cdot (-1) \]

\[ f'(x) = \frac{\pi}{2} - x - x = \frac{\pi}{2} - 2x \]

Step 2: Evaluate at \( x = -4 \)
Now substitute \( x = -4 \) into the derivative:

\[ f'(-4) = \frac{\pi}{2} - 2(-4) = \frac{\pi}{2} + 8 \]

Simplify:

\[ f'(-4) = \frac{\pi}{2} + \frac{16}{2} = \frac{\pi + 16}{2} \]

Thus, the value of \( f'(-4) \) is:

\[ f'(-4) = \frac{16 + \pi}{2} \]Thus, the correct answer is option (B), \( \frac{16 + \pi}{2} \).