1. **State the problem parts:** (a) State the three conditions for continuity of function $f(x)$ at $x = x_0$. (b) Determine if the given piecewise function $$ f(x) = \begin{cases} x^2, & x < 2 \\ 4, & x = 2 \\ -x + 6, & x > 2 \end{cases} $$ is continuous at $x=2$. 2. **Three conditions for continuity at $x = x_0$ are:** 1. The function is defined at $x_0$: $f(x_0)$ exists. 2. The limit of $f(x)$ as $x$ approaches $x_0$ exists: $\lim_{x \to x_0} f(x)$ exists. 3. The limit equals the function value: $\lim_{x \to x_0} f(x) = f(x_0)$. 3. **Check continuity at $x=2$ for given function:** - Calculate $f(2)$: $$ f(2) = 4 $$ - Calculate left-hand limit: $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4 $$ - Calculate right-hand limit: $$ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-x + 6) = -2 + 6 = 4 $$ - Since left-hand limit = right-hand limit = $f(2) = 4$, the function is continuous at $x=2$. 4. **Next problem:** (a) Find $f'(x)$ from first principles where $f(x) = 3x^2 + 2x + 3$. Definition of derivative using first principles: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ Evaluate numerator: $$ f(x+h) = 3(x+h)^2 + 2(x+h) + 3 = 3(x^2 + 2xh + h^2) + 2x + 2h + 3 = 3x^2 + 6xh + 3h^2 + 2x + 2h + 3 $$ So, $$ f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 + 2x + 2h + 3) - (3x^2 + 2x + 3) = 6xh + 3h^2 + 2h $$ Divide by $h$: $$ \frac{f(x+h) - f(x)}{h} = 6x + 3h + 2 $$ Take the limit as $h \to 0$: $$ f'(x) = \lim_{h \to 0} (6x + 3h + 2) = 6x + 2 $$ 5. **(b) Find $\frac{dy}{dx}$ given $y^2 + 2xy - x^4 = 0$ (implicit differentiation):** Differentiate both sides with respect to $x$: $$ \frac{d}{dx}(y^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(x^4) = 0 $$ Recall $y$ depends on $x$, use chain and product rules: $$ 2y \frac{dy}{dx} + 2\left(x \frac{dy}{dx} + y \right) - 4x^3 = 0 $$ Simplify: $$ 2y \frac{dy}{dx} + 2x \frac{dy}{dx} + 2y - 4x^3 = 0 $$ Group terms with $\frac{dy}{dx}$: $$ (2y + 2x) \frac{dy}{dx} = 4x^3 - 2y $$ Solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{4x^3 - 2y}{2y + 2x} = \frac{4x^3 - 2y}{2(y + x)} $$ 6. **(c) Evaluate the integral** $$ \int_0^{\pi} x \sin x \, dx $$ Use integration by parts: Let $$ u = x \implies du = dx $$ $$ dv = \sin x dx \implies v = -\cos x $$ Apply formula: $$ \int u \, dv = uv - \int v \, du $$ Thus $$ \int_0^{\pi} x \sin x \, dx = \left[-x \cos x\right]_0^{\pi} + \int_0^{\pi} \cos x \, dx $$ Evaluate boundary term: $$ -\pi \cos \pi + 0 = -\pi (-1) = \pi $$ Evaluate integral: $$ \int_0^{\pi} \cos x \, dx = \sin x \big|_0^{\pi} = \sin \pi - \sin 0 = 0 $$ So the integral equals $$ \pi + 0 = \pi $$ **Final answers:** (a) The function is continuous at $x=2$ because it satisfies all three conditions. (b) $f'(x) = 6x + 2$ (c) $\frac{dy}{dx} = \frac{4x^3 - 2y}{2(x+y)}$ (d) $\int_0^{\pi} x \sin x \, dx = \pi$