1. **Problem 1:** Determine if the function $$ f(x) = \begin{cases} \frac{\sin x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases} $$ is continuous at $x=0$. **Step 1:** Check the limit of $f(x)$ as $x \to 0$. Because $f(x) = \frac{\sin x}{|x|}$ for $x \neq 0$, consider the left-hand limit (as $x \to 0^-$) and the right-hand limit (as $x \to 0^+$): - For $x \to 0^+$, $|x| = x$, so $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin x}{x} = 1.$$ - For $x \to 0^-$, $|x| = -x$, so $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{-x} = - \lim_{x \to 0^-} \frac{\sin x}{x} = -1.$$ **Step 2:** Since the left-hand and right-hand limits as $x \to 0$ are not equal ($-1 \neq 1$), the limit $\lim_{x \to 0} f(x)$ does not exist. **Step 3:** For continuity at $x=0$, we need $\lim_{x \to 0} f(x) = f(0) = 1$. Since the limit does not exist, $f$ is **not continuous at $x=0$**. --- 2. **Problem 2:** Determine whether the function $$ f(x) = \ln(\sin x + 1) $$ is continuous on the interval $(0,\pi)$ and find $f'(x)$ at $x = \frac{\pi}{4}$. **Step 1:** Check the domain and continuity. - The argument of the logarithm must be positive: $\sin x + 1 > 0$. - Since $\sin x$ on $(0,\pi)$ lies in $(0,1]$, $\sin x + 1$ is always in $(1,2]$. - Therefore, $f(x)$ is defined and continuous on $(0,\pi)$. **Step 2:** Differentiate $f(x)$ using the chain rule: $$ f'(x) = \frac{1}{\sin x + 1} \cdot \cos x. $$ **Step 3:** Evaluate at $x = \frac{\pi}{4}$: $$ f'\left(\frac{\pi}{4}\right) = \frac{\cos \frac{\pi}{4}}{\sin \frac{\pi}{4} + 1} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} + 1} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}. $$ This can be simplified by finding a common denominator: $$ f'\left(\frac{\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{\frac{2}{2} + \frac{\sqrt{2}}{2}} = \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{2}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}}. $$ Multiply numerator and denominator by the conjugate to rationalize: $$ f'\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{\sqrt{2}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2} = \frac{2\sqrt{2} - 2}{4 - 2} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1. $$ **Final answer:** $f'(\frac{\pi}{4}) = \sqrt{2} - 1$. --- 3. **Problem 3:** For $$ f(x) = \sin x + e^{2x}, $$ find the instantaneous rate of change at $x=\frac{\pi}{4}$ and the equation of the tangent line at that point. **Step 1:** Differentiate $f(x)$: $$ f'(x) = \cos x + 2e^{2x}. $$ **Step 2:** Evaluate the derivative at $x=\frac{\pi}{4}$: $$ f'\left(\frac{\pi}{4}\right) = \cos \frac{\pi}{4} + 2e^{\frac{\pi}{2}} = \frac{\sqrt{2}}{2} + 2e^{\frac{\pi}{2}}. $$ **Step 3:** Find the point on the function at $x = \frac{\pi}{4}$: $$ f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} + e^{\frac{\pi}{2}} = \frac{\sqrt{2}}{2} + e^{\frac{\pi}{2}}. $$ **Step 4:** Equation of tangent line at $x_0 = \frac{\pi}{4}$ is $$ y - f\left(\frac{\pi}{4}\right) = f'\left(\frac{\pi}{4}\right)(x - \frac{\pi}{4}). $$ Substitute values: $$ y - \left(\frac{\sqrt{2}}{2} + e^{\frac{\pi}{2}}\right) = \left(\frac{\sqrt{2}}{2} + 2e^{\frac{\pi}{2}}\right) \left(x - \frac{\pi}{4}\right). $$ --- 4. **Problem 4:** (a) Explain difference between continuity and differentiability, provide example. **Step 1:** Continuity means the function has no breaks, jumps, or holes at a point (the limit of $f(x)$ as $x$ approaches that point equals the function value there). **Step 2:** Differentiability means the function has a defined slope (derivative) at that point; the function must be smooth without sharp corners. **Step 3:** A function can be continuous but not differentiable at a point. For example, the absolute value function $f(x) = |x|$ is continuous everywhere but not differentiable at $x=0$ because of the cusp (sharp corner). (b) Relation between rate of change and tangent line slope, behavior at nondifferentiable points. **Step 1:** The derivative of a function at a point gives the instantaneous rate of change and the slope of the tangent line. **Step 2:** At points where the function is not differentiable (like corners or cusps), the tangent line either does not exist or is not uniquely defined because the slope changes abruptly. **Step 3:** This means there is no single well-defined instantaneous rate of change at those points even though the function may be continuous there.