1. **Problem 1: Finding (f o f o f)(1) given the curve y = f(x).** The graph shows the function $f(x)$ with key points: $f(1) = 2$ (from the given approximate coordinates near x=1), $f(2) = -1$ (interpreting the curve crossing near these points), and $f(-1) = 1$ assuming symmetry or by reading the curve. We need to calculate $(f \,o\\, f \,o\\, f)(1) = f(f(f(1)))$. Step-by-step: 1. Calculate $f(1)$ - from the graph, at $x=1$, $f(1) \approx 2$. 2. Calculate $f(f(1)) = f(2)$ - looking at $x=2$, $f(2) \approx -1$. 3. Calculate $f(f(f(1))) = f(-1)$ - at $x=-1$, $f(-1) \approx 1$. Therefore, $(f o f o f)(1) = 1$. Answer: (c) 1 2. **Problem 2: Domain of the function $f(x) = \sqrt{x - 2} - \sqrt{7 - x}$** The domain is where the expressions under the square roots are non-negative: $$x - 2 \geq 0 \Rightarrow x \geq 2$$ $$7 - x \geq 0 \Rightarrow x \leq 7$$ So, the domain is: $$[2, 7]$$ Answer: (a) [2, 7] 3. **Problem 3: Domain of the function $(f \circ g)(x)$, where** $$f(x) = x^{2} - 4, \quad g(x) = x + 1$$ The composition is: $$(f \circ g)(x) = f(g(x)) = (x + 1)^{2} - 4 = x^{2} + 2x + 1 - 4 = x^{2} + 2x - 3$$ Since this is a polynomial, its domain is all real numbers, $ $$\mathbb{R}$$ Answer: (c) $\mathbb{R}$