1. **State the problem:** We have three functions: $$f(x) = \frac{3}{x+2}, \quad g(x) = 4x - 5, \quad h(x) = x^2 + 1.$$ We need to find: (i) The value of $x$ for which $f(x)$ is undefined. (ii) The values of: a) $g\left(\frac{1}{4}\right)$ b) $h(-3)$ c) $f(f'(0))$ 2. **Find when $f(x)$ is undefined:** $f(x)$ is undefined when the denominator is zero: $$x + 2 = 0$$ $$x = -2$$ 3. **Calculate $g\left(\frac{1}{4}\right)$:** Substitute $x = \frac{1}{4}$ into $g(x)$: $$g\left(\frac{1}{4}\right) = 4 \times \frac{1}{4} - 5 = 1 - 5 = -4$$ 4. **Calculate $h(-3)$:** Substitute $x = -3$ into $h(x)$: $$h(-3) = (-3)^2 + 1 = 9 + 1 = 10$$ 5. **Calculate $f(f'(0))$:** First, find the derivative $f'(x)$ of $f(x) = \frac{3}{x+2}$: Rewrite $f(x)$ as: $$f(x) = 3(x+2)^{-1}$$ Using the power rule and chain rule: $$f'(x) = 3 \times (-1)(x+2)^{-2} \times 1 = -\frac{3}{(x+2)^2}$$ Evaluate $f'(0)$: $$f'(0) = -\frac{3}{(0+2)^2} = -\frac{3}{4}$$ Now find $f\left(f'(0)\right) = f\left(-\frac{3}{4}\right)$: $$f\left(-\frac{3}{4}\right) = \frac{3}{-\frac{3}{4} + 2} = \frac{3}{-\frac{3}{4} + \frac{8}{4}} = \frac{3}{\frac{5}{4}} = 3 \times \frac{4}{5} = \frac{12}{5}$$ **Final answers:** (i) $x = -2$ (ii) a) $-4$ b) $10$ c) $\frac{12}{5}$