Let's tackle the problem step by step: ### 1. The function: \[ C(x) = \int_0^x \cos\left(\frac{1}{2}\pi t^2\right) dt \] --- #### **a) Find the derivative of \( C(x) \)** By the **Fundamental Theorem of Calculus**: \[ C'(x) = \frac{d}{dx} \left( \int_0^x \cos\left(\frac{1}{2}\pi t^2\right) dt \right) = \cos\left(\frac{1}{2}\pi x^2\right) \] --- #### **b) On what intervals is the function \( C \) increasing?** A function increases when its derivative is positive. \[ C'(x) = \cos\left(\frac{1}{2}\pi x^2\right) > 0 \] So, \( C(x) \) is **increasing** where: \[ \frac{1}{2}\pi x^2 \in (2\pi n - \frac{\pi}{2}, 2\pi n + \frac{\pi}{2}) \quad \text{for integers } n \] This simplifies to: \[ 2\pi n - \frac{\pi}{2} < \frac{1}{2}\pi x^2 < 2\pi n + \frac{\pi}{2} \] Divide both sides by \( \frac{1}{2}\pi \): \[ 4n - 1 < x^2 < 4n + 1 \] So, \[ x \in (-\sqrt{4n + 1}, -\sqrt{4n - 1}) \cup (\sqrt{4n - 1}, \sqrt{4n + 1}) \] for all \( n \geq 1 \) (since \( x^2 > 0 \)), and for \( n = 0 \), \( x \in (-1,1) \). --- #### **c) On what intervals is the function \( C \) decreasing?** A function decreases when its derivative is negative: \[ C'(x) = \cos\left(\frac{1}{2}\pi x^2\right) < 0 \] So, \( C(x) \) is **decreasing** where: \[ \frac{1}{2}\pi x^2 \in (2\pi n + \frac{\pi}{2}, 2\pi n + \frac{3\pi}{2}) \] This simplifies to: \[ 4n + 1 < x^2 < 4n + 3 \] So, \[ x \in (-\sqrt{4n + 3}, -\sqrt{4n + 1}) \cup (\sqrt{4n + 1}, \sqrt{4n + 3}) \] for all \( n \geq 0 \). --- ### **Summary Table** | Part | Answer | |------|--------| | (a) | \( C'(x) = \cos\left(\frac{1}{2}\pi x^2\right) \) | | (b) | \( C(x) \) is **increasing** on \( (-\sqrt{4n + 1}, -\sqrt{4n - 1}) \cup (\sqrt{4n - 1}, \sqrt{4n + 1}) \), \( n = 1,2,3,... \) and \( (-1,1) \) for \( n = 0 \) | | (c) | \( C(x) \) is **decreasing** on \( (-\sqrt{4n + 3}, -\sqrt{4n + 1}) \cup (\sqrt{4n + 1}, \sqrt{4n + 3}) \), \( n = 0,1,2,... \) | Let me know if you'd like to see this with an explicit computation for the first few intervals!

Let's carefully work through each part of the problem with detailed explanations and calculations. --- ### Given Function: \[ C(x) = \int_0^x \cos\left(\frac{1}{2}\pi t^2\right) dt \] --- ## Part (a): Find the derivative \( C'(x) \) ### Step 1: Recall the Fundamental Theorem of Calculus - If \( C(x) = \int_a^x f(t) dt \), then \( C'(x) = f(x) \). ### Step 2: Identify \( f(t) \) - Here, \( f(t) = \cos\left(\frac{1}{2}\pi t^2\right) \). ### Step 3: Write the derivative \[ C'(x) = \cos\left(\frac{1}{2}\pi x^2\right) \] **Final answer for part (a):** \[ \boxed{ \boxed{ C'(x) = \cos\left(\frac{1}{2}\pi x^2\right) } } \] --- ## Part (b): Find intervals where \( C \) is increasing ### Step 1: Recall that \( C \) is increasing where \( C'(x) > 0 \). \[ \Rightarrow \cos\left(\frac{1}{2}\pi x^2\right) > 0 \] ### Step 2: Find where cosine is positive - \( \cos \theta > 0 \) in intervals: \[ \theta \in (2\pi n - \frac{\pi}{2}, 2\pi n + \frac{\pi}{2}) \quad \text{for all integers } n \] but more precisely, cosine is positive in the intervals: \[ \theta \in (-\frac{\pi}{2} + 2\pi n, \frac{\pi}{2} + 2\pi n) \] ### Step 3: Set \( \theta = \frac{1}{2}\pi x^2 \) \[ \frac{1}{2}\pi x^2 \in \left( 2\pi n - \frac{\pi}{2}, 2\pi n + \frac{\pi}{2} \right) \] ### Step 4: Solve for \( x^2 \) Multiply all parts by 2: \[ \pi x^2 \in \left( 4\pi n - \pi, 4\pi n + \pi \right) \] Divide by \( \pi \): \[ x^2 \in \left( 4 n - 1, 4 n + 1 \right) \] ### Step 5: Find \( x \) \[ x \in \left( -\sqrt{4 n + 1}, -\sqrt{4 n - 1} \right) \cup \left( \sqrt{4 n - 1}, \sqrt{4 n + 1} \right) \] - For \( n=0 \): \[ x^2 \in (-1, 1) \Rightarrow x \in (-1, 1) \] - For \( n \geq 1 \): \[ x \in \left( -\sqrt{4 n + 1}, -\sqrt{4 n - 1} \right) \cup \left( \sqrt{4 n - 1}, \sqrt{4 n + 1} \right) \] --- ### **Answer for part (b):** \[ \boxed{ \text{C is increasing on} \quad (-1, 1) \quad \text{and} \quad \bigcup_{n=1}^\infty \left[ \left( -\sqrt{4n+1}, -\sqrt{4n-1} \right) \cup \left( \sqrt{4n-1}, \sqrt{4n+1} \right) \right] } \] --- ## Part (c): Find intervals where \( C \) is decreasing ### Step 1: \( C \) is decreasing where \( C'(x) < 0 \). \[ \Rightarrow \cos\left(\frac{1}{2}\pi x^2\right) < 0 \] ### Step 2: Determine where cosine is negative - Cosine is negative in intervals: \[ \theta \in \left( \pi/2 + 2\pi n, 3\pi/2 + 2\pi n \right), \quad n \in \mathbb{Z} \] ### Step 3: Set \( \theta = \frac{1}{2}\pi x^2 \) \[ \frac{1}{2}\pi x^2 \in \left( \frac{\pi}{2} + 2\pi n, \frac{3\pi}{2} + 2\pi n \right) \] ### Step 4: Solve for \( x^2 \) Multiply all parts by 2: \[ \pi x^2 \in \left( \pi + 4\pi n, 3\pi + 4\pi n \right) \] Divide by \( \pi \): \[ x^2 \in \left( 1 + 4 n, 3 + 4 n \right) \] ### Step 5: Find \( x \) \[ x \in \left( -\sqrt{3 + 4 n}, -\sqrt{1 + 4 n} \right) \cup \left( \sqrt{1 + 4 n}, \sqrt{3 + 4 n} \right) \] - For \( n=0 \): \[ x^2 \in (1, 3) \Rightarrow x \in (-\sqrt{3}, -1) \cup (1, \sqrt{3}) \] - For \( n \geq 1 \): \[ x \in \left( -\sqrt{4 n + 3}, -\sqrt{4 n + 1} \right) \cup \left( \sqrt{4 n + 1}, \sqrt{4 n + 3} \right) \] --- ### **Answer for part (c):** \[ \boxed{ \text{C is decreasing on} \quad \bigcup_{n=0}^\infty \left[ \left( -\sqrt{4 n + 3}, -\sqrt{4 n + 1} \right) \cup \left( \sqrt{4 n + 1}, \sqrt{4 n + 3} \right) \right] } \] --- # **Summary:** | Part | Derivative \( C'(x) \) | Intervals of Increase | Intervals of Decrease | |--------|------------------------------|--------------------------------|---------------------------------| | (a) | \( \cos\left(\frac{1}{2} \pi x^2 \right) \) | \( (-1, 1) \), and for \( n \geq 1 \), \( (\pm \sqrt{4n-1}, \pm \sqrt{4n+1}) \) | For \( n \geq 0 \), \( (\pm \sqrt{4n+1}, \pm \sqrt{4n+3}) \) | Let me know if you'd like a specific numerical example or further clarification!