Let's go **step-by-step** for each function, focusing on the **2π-periodic extension**, and then analyze **continuity** and **differentiability**. --- ## General Idea Given a function \( f(x) \) defined on \( [-\pi, \pi] \), its **2π-periodic extension** is: \[ f_{per}(x) = f(x - 2\pi n) \text{ where } n \in \mathbb{Z} \text{ so that } x - 2\pi n \in [-\pi, \pi]. \] We study: - **Continuity** at the endpoints \( x = -\pi \) and \( x = \pi \). - **Differentiability** at the same points. - **Elsewhere** the function is as smooth as the original. --- ### (a) \( x^2 \) On \( [-\pi, \pi] \): \( f(x) = x^2 \) - **At \( x = -\pi \):** \( f(-\pi) = \pi^2 \) - **At \( x = \pi \):** \( f(\pi) = \pi^2 \) So, values at endpoints match. - **Continuity:** Yes, it's continuous. - **Differentiability:** \( f'(x) = 2x \), so \( f'(-\pi) = -2\pi \), \( f'(\pi) = 2\pi \); these are **not equal**. So, **not differentiable** at the endpoints. --- ### (b) \( (x^2 - \pi^2)^2 \) On \( [-\pi, \pi] \): \( f(x) = (x^2 - \pi^2)^2 \) - **At \( x = -\pi \) and \( x = \pi \):** \( f = 0 \) - **Continuity:** Yes, endpoints match. - **Differentiability:** \( f'(x) = 4x(x^2 - \pi^2) \) - \( f'(-\pi) = 4(-\pi)(0) = 0 \) - \( f'(\pi) = 4(\pi)(0) = 0 \) - Endpoints match! **Differentiable**. --- ### (c) \( e^x \) - **At \( x = -\pi \):** \( e^{-\pi} \) - **At \( x = \pi \):** \( e^{\pi} \) - **Endpoints do NOT match**. - **Continuity:** **Not continuous**. - **Differentiability:** Not differentiable (not even continuous). --- ### (d) \( e^{-|x|} \) - Even function, peaks at \( x = 0 \), decays to \( e^{-\pi} \) at endpoints. - **At both \( x = \pm\pi \):** \( e^{-\pi} \) - **Continuity:** Yes, endpoints match. - **Differentiability:** Check left/right derivatives at \( x = \pi \): - \( f'(x) = -e^{-|x|} \) for \( x > 0 \), \( e^{-|x|} \) for \( x < 0 \). - At endpoint, periodic extension jumps in slope. - Compute: - \( \lim_{x \to \pi^-} f'(x) = -e^{-\pi} \) - \( \lim_{x \to -\pi^+} f'(x) = e^{-\pi} \) - **Not differentiable** at endpoints. --- ### (e) \( \sinh x \) - **Odd function**, but - \( \sinh(-\pi) = -\sinh(\pi) \) - **Endpoints DO NOT match**. - **Continuity:** Not continuous. --- ### (f) \( 1 + \cos^2 x \) - \( \cos^2 x \) is \( 2\pi \) periodic. - So, the **function is already periodic**. - **Continuous:** Yes. - **Differentiable:** Yes, everywhere. --- ### (g) \( \sin \left(\frac{1}{2}\pi x\right) \) - Period of this function is \( T = 4 \). - On \([-\pi, \pi]\), endpoints: - \( x = -\pi: \sin(-\frac{1}{2}\pi^2) \) - \( x = \pi: \sin(\frac{1}{2}\pi^2) \) - These are **not equal**. - Not continuous. --- ### (h) \( \frac{1}{x} \) - Has a singularity at \( x = 0 \). - Not even defined on all of \([-\pi, \pi]\). - So **not continuous**. --- ### (i) \( \frac{1}{1 + x^2} \) - Smooth, even function, values at endpoints: - \( x = \pm\pi: \frac{1}{1+\pi^2} \) - **Continuous:** Yes. - **Differentiable:** Yes, as derivatives at endpoints are equal (even function, smooth). --- ## **Summary Table** | Function | Continuous? | Differentiable? | |--------------------------|-------------|-----------------| | (a) \(x^2\) | Yes | No | | (b) \((x^2-\pi^2)^2\) | Yes | Yes | | (c) \(e^x\) | No | No | | (d) \(e^{-|x|}\) | Yes | No | | (e) \(\sinh x\) | No | No | | (f) \(1+\cos^2 x\) | Yes | Yes | | (g) \(\sin(\frac12\pi x)\)| No | No | | (h) \(1/x\) | No | No | | (i) \(\frac{1}{1+x^2}\) | Yes | Yes | --- ### **Final Answers** - **Continuous and differentiable:** (b), (f), (i) - **Continuous but not differentiable:** (a), (d) - **Not continuous:** (c), (e), (g), (h) If you need **graphs** or further step-by-step calculations for any part, please specify!

Certainly! Let's go through each function carefully, explaining how to construct its 2π-periodic extension, analyze its continuity and differentiability, and then provide the final conclusion. --- ## General Approach Given a function \(f(x)\) defined on \([- \pi, \pi]\), its **2π-periodic extension** is: \[ f_{per}(x) = f(x - 2\pi n) \quad \text{where} \quad n \in \mathbb{Z} \text{ is chosen so that } x - 2\pi n \in [-\pi, \pi]. \] This essentially "wraps" the original function periodically every \(2\pi\). To analyze properties like continuity and differentiability, we focus on the behavior at the **period boundaries** \(x = -\pi\) and \(x = \pi\). --- ## (a) \(f(x) = x^2\) ### On \([- \pi, \pi]\): - \(f(-\pi) = (\,-\pi\,)^2 = \pi^2\) - \(f(\pi) = \pi^2\) ### Step-by-step: - The function is **smooth** (polynomial), so **continuous** and **differentiable** on \((- \pi, \pi)\). - At the endpoints: \[ f(-\pi) = f(\pi) = \pi^2, \] so the periodic extension "matches up" in terms of value. - **Continuity of the extension:** The function is continuous across the boundary points because the values match. - **Differentiability at the boundary:** \[ f'(x) = 2x, \] so: \[ f'(-\pi) = -2\pi, \quad f'(\pi) = 2\pi. \] Since these are **not equal**, the periodic extension has a **discontinuous derivative** at \(x = \pm \pi\). ### **Summary:** - **Periodic extension is continuous everywhere**. - **Not differentiable at the boundary points**. --- ## (b) \(f(x) = (x^2 - \pi^2)^2\) ### On \([- \pi, \pi]\): - Values at the endpoints: \[ f(-\pi) = ((-\pi)^2 - \pi^2)^2 = ( \pi^2 - \pi^2)^2 = 0, \] \[ f(\pi) = (\pi^2 - \pi^2)^2 = 0. \] - The function is smooth inside the interval. ### Derivative: \[ f'(x) = 2(x^2 - \pi^2) \times 2x = 4x(x^2 - \pi^2). \] At the endpoints: \[ f'(-\pi) = 4(-\pi)(\pi^2 - \pi^2) = 0, \] \[ f'(\pi) = 4(\pi)(\pi^2 - \pi^2) = 0, \] which match. ### **Conclusion:** - The function's value matches at the endpoints, so the extension is **continuous**. - The derivatives match at the endpoints, so the extension is **differentiable** there. --- ## (c) \(f(x) = e^x\) ### On \([- \pi, \pi]\): - Endpoints: \[ f(-\pi) = e^{- \pi}, \quad f(\pi) = e^{\pi}. \] - Since \(e^{-\pi} \neq e^{\pi}\), the periodic extension **jumps** at the boundary points. ### **Conclusion:** - The extension **is not continuous** at \(x = \pm \pi\). - Therefore, it is **not differentiable** there. --- ## (d) \(f(x) = e^{-|x|}\) ### On \([- \pi, \pi]\): - Values: \[ f(-\pi) = e^{-\pi}, \quad f(\pi) = e^{-\pi}. \] - The values match, so the extension is **continuous** at the boundary points. ### Derivative: \[ f'(x) = -\operatorname{sgn}(x) e^{-|x|}, \] where \(\operatorname{sgn}(x)\) is the sign function: \[ \operatorname{sgn}(x) = \begin{cases} -1, & x < 0, \\ 0, & x = 0, \\ 1, & x > 0. \end{cases} \] - At \(x = \pm \pi\), the derivatives from the left and right are: \[ \lim_{x \to \pi^-} f'(x) = -e^{-\pi}, \quad \lim_{x \to -\pi^+} f'(x) = e^{-\pi}. \] - The derivatives **do not match** at the boundary points, so the extension **is not differentiable** there. --- ## (e) \(f(x) = \sinh x\) ### On \([- \pi, \pi]\): - \(\sinh x\) is an odd and smooth function. - Endpoints: \[ \sinh(-\pi) = -\sinh \pi, \] \[ \sinh(\pi) = \sinh \pi. \] - Values are **not equal** in magnitude but opposites, so the periodic extension causes a jump at the boundary points. ### **Conclusion:** - **Not continuous** at \(x = \pm \pi\). --- ## (f) \(f(x) = 1 + \cos^2 x\) - \(\cos^2 x\) has period \( \pi \), but since we're extending periodically with period \(2\pi\), the function is **already periodic**. - On \([- \pi, \pi]\): \[ f(-\pi) = 1 + \cos^2(-\pi) = 1 + \cos^2 \pi = 1 + 1 = 2, \] \[ f(\pi) = 1 + \cos^2 \pi = 2. \] - Values match, and the function is smooth everywhere, so the periodic extension is: - **Continuous**, and - **Differentiable** everywhere. --- ## (g) \(f(x) = \sin \left(\frac{\pi}{2} x \right)\) - Period of \(\sin \left( \frac{\pi}{2} x \right)\) is: \[ T = \frac{2\pi}{\frac{\pi}{2}} = 4, \] so the function repeats every 4 units. - On \([- \pi, \pi]\): \[ f(-\pi) = \sin \left( -\frac{\pi}{2} \pi \right) = \sin(-\frac{\pi^2}{2}), \] \[ f(\pi) = \sin \left( \frac{\pi}{2} \pi \right) = \sin(\frac{\pi^2}{2}), \] which **are not equal** in general. - Since the period is 4, extending periodically with period \(2\pi\) (which is approximately 6.283) causes mismatches at the boundary points. - **Conclusion:** - The function **is not continuous** at the boundary points when extended periodically. --- ## (h) \(f(x) = \frac{1}{x}\) - Has a **singularity at \(x=0\)**. - Cannot be extended periodically over \([- \pi, \pi]\) without issues. - **Not continuous** on the interval. --- ## (i) \(f(x) = \frac{1}{1 + x^2}\) - Smooth, with period \(2\pi\) (or any period, if extended). - On \([- \pi, \pi]\): \[ f(-\pi) = \frac{1}{1 + \pi^2}, \quad f(\pi) = \frac{1}{1 + \pi^2}. \] - Values match; derivatives: \[ f'(x) = -\frac{2x}{(1 + x^2)^2}, \] which also match at endpoints. - **Conclusion:** - The periodic extension is **both continuous and differentiable** everywhere. --- ## **Summary of Results** | Function | Is the periodic extension **continuous**? | Is it **differentiable**? | |---|---|---| | (a) \(x^2\) | Yes | **No** (due to derivative jump) | | (b) \((x^2 - \pi^2)^2\) | Yes | Yes | | (c) \(e^x\) | No | No | | (d) \(e^{-|x|}\) | Yes | No | | (e) \(\sinh x\) | No | No | | (f) \(1 + \cos^2 x\) | Yes | Yes | | (g) \(\sin(\frac{\pi}{2} x)\) | No | No | | (h) \(1/x\) | No | No | | (i) \(\frac{1}{1 + x^2}\) | Yes | Yes | --- ## Final note: - The key is checking endpoint values and derivatives. - Continuous extensions require matching boundary values. - Differentiability requires matching derivatives at boundary points. If you'd like, I can also help you draw these periodic extensions or explore their graphs!

Certainly! Here are the final answers for each part regarding the 2π-periodic extension, its continuity, and differentiability: 1. **\(f(x) = x^2\)** - **Continuous:** Yes - **Differentiable:** No (derivative jumps at \(x = \pm \pi\)) 2. **\(f(x) = (x^2 - \pi^2)^2\)** - **Continuous:** Yes - **Differentiable:** Yes 3. **\(f(x) = e^x\)** - **Continuous:** No - **Differentiable:** No 4. **\(f(x) = e^{-|x|}\)** - **Continuous:** Yes - **Differentiable:** No (non-smooth at boundary points) 5. **\(f(x) = \sinh x\)** - **Continuous:** No - **Differentiable:** No 6. **\(f(x) = 1 + \cos^2 x\)** - **Continuous:** Yes - **Differentiable:** Yes 7. **\(f(x) = \sin \left(\frac{\pi}{2} x \right)\)** - **Continuous:** No - **Differentiable:** No 8. **\(f(x) = \frac{1}{x}\)** - **Continuous:** No (singularity at 0) - **Differentiable:** No 9. **\(f(x) = \frac{1}{1 + x^2}\)** - **Continuous:** Yes - **Differentiable:** Yes These are the summarized final answers for each function's periodic extension regarding their continuity and differentiability.