1. **Define continuity of a function $f(x)$ at a point $x=c$: ** A function $f(x)$ is continuous at $x=c$ if the following three conditions hold: 1. $f(c)$ is defined. 2. The limit $\lim_{x \to c} f(x)$ exists. 3. $\lim_{x \to c} f(x) = f(c)$. This means the function has no breaks, jumps, or holes at $x=c$. 2. **Differentiate using the product rule:** The product rule states: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ (i) $h(x) = (7x + 4x^2)(9 - 9x)$ - Let $u = 7x + 4x^2$, $v = 9 - 9x$ - Compute derivatives: $u' = 7 + 8x$ $v' = -9$ - Apply product rule: $$h'(x) = (7 + 8x)(9 - 9x) + (7x + 4x^2)(-9)$$ - Simplify: $$= (7 + 8x)(9 - 9x) - 9(7x + 4x^2)$$ $$= 63 - 63x + 72x - 72x^2 - 63x - 36x^2$$ $$= 63 + 9x - 108x^2$$ (ii) $y = 8x^4 \cos x$ - Let $u = 8x^4$, $v = \cos x$ - Derivatives: $u' = 32x^3$ $v' = -\sin x$ - Product rule: $$y' = 32x^3 \cos x + 8x^4 (-\sin x) = 32x^3 \cos x - 8x^4 \sin x$$ (iii) $f(x) = x^5 \sqrt{1 + 4x^3} = x^5 (1 + 4x^3)^{1/2}$ - Let $u = x^5$, $v = (1 + 4x^3)^{1/2}$ - Derivatives: $u' = 5x^4$ $$v' = \frac{1}{2}(1 + 4x^3)^{-1/2} \cdot 12x^2 = \frac{6x^2}{\sqrt{1 + 4x^3}}$$ - Product rule: $$f' = 5x^4 (1 + 4x^3)^{1/2} + x^5 \cdot \frac{6x^2}{\sqrt{1 + 4x^3}} = 5x^4 \sqrt{1 + 4x^3} + \frac{6x^7}{\sqrt{1 + 4x^3}}$$ 3. **Discuss continuity of the piecewise function:** $$f(x) = \begin{cases} 2x - 2 & x < 3 \\ 3x & 3 \leq x \leq 5 \\ x^2 + x & x > 5 \end{cases}$$ - Check continuity at $x=3$: - Left limit: $\lim_{x \to 3^-} f(x) = 2(3) - 2 = 4$ - Right limit: $\lim_{x \to 3^+} f(x) = 3(3) = 9$ - Since left and right limits differ, $f$ is discontinuous at $x=3$. - Check continuity at $x=5$: - Left limit: $\lim_{x \to 5^-} f(x) = 3(5) = 15$ - Right limit: $\lim_{x \to 5^+} f(x) = 5^2 + 5 = 30$ - Limits differ, so $f$ is discontinuous at $x=5$. - For other $x$, each piece is continuous. 4. **Find $y''$ given $y = 3x^2 + 4e^x + \cos x$:** - First derivative: $$y' = 6x + 4e^x - \sin x$$ - Second derivative: $$y'' = 6 + 4e^x - \cos x$$ **Final answers:** (i) $h'(x) = 63 + 9x - 108x^2$ (ii) $y' = 32x^3 \cos x - 8x^4 \sin x$ (iii) $f' = 5x^4 \sqrt{1 + 4x^3} + \frac{6x^7}{\sqrt{1 + 4x^3}}$ Continuity: Discontinuous at $x=3$ and $x=5$. $y'' = 6 + 4e^x - \cos x$