1. **Problem:** Given $f(x) = x^2 + 4x - 1$ and $g(x) = 5x - 2$, find: a) $(f \circ g)(x)$ b) $(g \circ f)(x)$ c) $(f \circ g)(5)$ **Step 1:** Recall that $(f \circ g)(x) = f(g(x))$ means substitute $g(x)$ into $f$. **Step 2:** Calculate $(f \circ g)(x)$: $$f(g(x)) = (5x - 2)^2 + 4(5x - 2) - 1$$ Expand: $$(5x - 2)^2 = 25x^2 - 20x + 4$$ So, $$f(g(x)) = 25x^2 - 20x + 4 + 20x - 8 - 1 = 25x^2 - 5$$ **Step 3:** Calculate $(g \circ f)(x) = g(f(x))$: $$g(f(x)) = 5(x^2 + 4x - 1) - 2 = 5x^2 + 20x - 5 - 2 = 5x^2 + 20x - 7$$ **Step 4:** Calculate $(f \circ g)(5)$: First find $g(5)$: $$g(5) = 5(5) - 2 = 25 - 2 = 23$$ Then, $$f(23) = 23^2 + 4(23) - 1 = 529 + 92 - 1 = 620$$ \boxed{(f \circ g)(x) = 25x^2 - 5, \quad (g \circ f)(x) = 5x^2 + 20x - 7, \quad (f \circ g)(5) = 620}\\ 2. **Problem:** $f(x) = \frac{5}{x+4}$, $g(x) = \frac{1}{x}$ **Step 1:** $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \frac{5}{\frac{1}{x} + 4}$ Simplify denominator: $$\frac{1}{x} + 4 = \frac{1 + 4x}{x}$$ So, $$f(g(x)) = \frac{5}{\frac{1 + 4x}{x}} = 5 \times \frac{x}{1 + 4x} = \frac{5x}{1 + 4x}$$ **Step 2:** $(g \circ f)(x) = g(f(x)) = g\left(\frac{5}{x+4}\right) = \frac{1}{\frac{5}{x+4}} = \frac{x+4}{5}$ **Step 3:** $(f \circ g)(5)$: $$g(5) = \frac{1}{5}$$ Then, $$f\left(\frac{1}{5}\right) = \frac{5}{\frac{1}{5} + 4} = \frac{5}{\frac{1}{5} + \frac{20}{5}} = \frac{5}{\frac{21}{5}} = 5 \times \frac{5}{21} = \frac{25}{21}$$ \boxed{(f \circ g)(x) = \frac{5x}{1 + 4x}, \quad (g \circ f)(x) = \frac{x+4}{5}, \quad (f \circ g)(5) = \frac{25}{21}}\\ 3. **Problem:** $f(x) = \frac{1}{x} + x$, $g(x) = \frac{1}{x}$ **Step 1:** $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} + \frac{1}{x} = x + \frac{1}{x}$ **Step 2:** $(g \circ f)(x) = g(f(x)) = \frac{1}{\frac{1}{x} + x} = \frac{1}{\frac{1+x^2}{x}} = \frac{x}{1 + x^2}$ **Step 3:** $(f \circ g)(5)$: $$g(5) = \frac{1}{5}$$ Then, $$f\left(\frac{1}{5}\right) = 5 + \frac{1}{5} = \frac{25}{5} + \frac{1}{5} = \frac{26}{5}$$ \boxed{(f \circ g)(x) = x + \frac{1}{x}, \quad (g \circ f)(x) = \frac{x}{1 + x^2}, \quad (f \circ g)(5) = \frac{26}{5}}\\ 4. **Problem:** $f(x) = 3x^2 + 1$, $g(x) = 4x - 3$ **Step 1:** $(f \circ g)(x) = f(g(x)) = 3(4x - 3)^2 + 1$ Expand: $$(4x - 3)^2 = 16x^2 - 24x + 9$$ So, $$f(g(x)) = 3(16x^2 - 24x + 9) + 1 = 48x^2 - 72x + 27 + 1 = 48x^2 - 72x + 28$$ **Step 2:** $(g \circ f)(x) = g(f(x)) = 4(3x^2 + 1) - 3 = 12x^2 + 4 - 3 = 12x^2 + 1$ **Step 3:** $(f \circ g)(5)$: $$g(5) = 4(5) - 3 = 20 - 3 = 17$$ Then, $$f(17) = 3(17)^2 + 1 = 3(289) + 1 = 867 + 1 = 868$$ \boxed{(f \circ g)(x) = 48x^2 - 72x + 28, \quad (g \circ f)(x) = 12x^2 + 1, \quad (f \circ g)(5) = 868}\\ 5. **Problem:** $f(x) = 2x^2 - x + 5$, $g(x) = x + 4$ **Step 1:** $(f \circ g)(x) = f(g(x)) = 2(x+4)^2 - (x+4) + 5$ Expand: $$(x+4)^2 = x^2 + 8x + 16$$ So, $$f(g(x)) = 2(x^2 + 8x + 16) - x - 4 + 5 = 2x^2 + 16x + 32 - x + 1 = 2x^2 + 15x + 33$$ **Step 2:** $(g \circ f)(x) = g(f(x)) = (2x^2 - x + 5) + 4 = 2x^2 - x + 9$ **Step 3:** $(f \circ g)(5)$: $$g(5) = 5 + 4 = 9$$ Then, $$f(9) = 2(9)^2 - 9 + 5 = 2(81) - 9 + 5 = 162 - 9 + 5 = 158$$ \boxed{(f \circ g)(x) = 2x^2 + 15x + 33, \quad (g \circ f)(x) = 2x^2 - x + 9, \quad (f \circ g)(5) = 158}\\ **Challenge Problem:** Find $(f \circ g \circ h)(5)$ where $h(x) = x^2 - 1$, $g(x) = x + 4$, $f(x) = 3x$ **Step 1:** Calculate $h(5)$: $$h(5) = 5^2 - 1 = 25 - 1 = 24$$ **Step 2:** Calculate $g(h(5)) = g(24)$: $$g(24) = 24 + 4 = 28$$ **Step 3:** Calculate $f(g(h(5))) = f(28)$: $$f(28) = 3(28) = 84$$ \boxed{(f \circ g \circ h)(5) = 84}