1. Composition of Functions: Given functions $f$ and $g$, find $(f \circ g)(x)$, $(g \circ f)(x)$, and $(f \circ g)(5)$ for each pair. **Important:** - $(f \circ g)(x) = f(g(x))$ means substitute $g(x)$ into $f$. - $(g \circ f)(x) = g(f(x))$ means substitute $f(x)$ into $g$. --- ### Problem 1 $f(x) = x^2 + 4x - 1$, $g(x) = 5x - 2$ 1. $(f \circ g)(x) = f(g(x)) = f(5x - 2) = (5x - 2)^2 + 4(5x - 2) - 1$ Expand: $$(5x - 2)^2 = 25x^2 - 20x + 4$$ So, $$(f \circ g)(x) = 25x^2 - 20x + 4 + 20x - 8 - 1 = 25x^2 - 5$$ 2. $(g \circ f)(x) = g(f(x)) = g(x^2 + 4x - 1) = 5(x^2 + 4x - 1) - 2 = 5x^2 + 20x - 5 - 2 = 5x^2 + 20x - 7$ 3. $(f \circ g)(5) = f(g(5)) = f(5(5) - 2) = f(25 - 2) = f(23) = 23^2 + 4(23) - 1 = 529 + 92 - 1 = 620$ \boxed{(f \circ g)(x) = 25x^2 - 5, \quad (g \circ f)(x) = 5x^2 + 20x - 7, \quad (f \circ g)(5) = 620}$ --- ### Problem 2 $f(x) = \frac{5}{x+4}$, $g(x) = \frac{1}{x}$ 1. $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \frac{5}{\frac{1}{x} + 4} = \frac{5}{\frac{1 + 4x}{x}} = 5 \cdot \frac{x}{1 + 4x} = \frac{5x}{1 + 4x}$ 2. $(g \circ f)(x) = g(f(x)) = g\left(\frac{5}{x+4}\right) = \frac{1}{\frac{5}{x+4}} = \frac{x+4}{5}$ 3. $(f \circ g)(5) = f(g(5)) = f\left(\frac{1}{5}\right) = \frac{5}{\frac{1}{5} + 4} = \frac{5}{\frac{1 + 20}{5}} = \frac{5}{\frac{21}{5}} = 5 \cdot \frac{5}{21} = \frac{25}{21}$ \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}}$ --- ### Problem 3 $f(x) = \frac{1}{x} + x$, $g(x) = \frac{1}{x}$ 1. $(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} + \frac{1}{x} = x + \frac{1}{x}$ 2. $(g \circ f)(x) = g(f(x)) = g\left(\frac{1}{x} + x\right) = \frac{1}{\frac{1}{x} + x} = \frac{1}{\frac{1 + x^2}{x}} = \frac{x}{1 + x^2}$ 3. $(f \circ g)(5) = f(g(5)) = f\left(\frac{1}{5}\right) = 5 + \frac{1}{5} = \frac{25}{5} + \frac{1}{5} = \frac{26}{5}$ \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}}$ --- ### Problem 4 $f(x) = 3x^2 + 1$, $g(x) = 4x - 3$ 1. $(f \circ g)(x) = f(g(x)) = f(4x - 3) = 3(4x - 3)^2 + 1$ Expand: $$(4x - 3)^2 = 16x^2 - 24x + 9$$ So, $$(f \circ g)(x) = 3(16x^2 - 24x + 9) + 1 = 48x^2 - 72x + 27 + 1 = 48x^2 - 72x + 28$$ 2. $(g \circ f)(x) = g(f(x)) = g(3x^2 + 1) = 4(3x^2 + 1) - 3 = 12x^2 + 4 - 3 = 12x^2 + 1$ 3. $(f \circ g)(5) = f(g(5)) = f(4(5) - 3) = f(20 - 3) = f(17) = 3(17)^2 + 1 = 3(289) + 1 = 867 + 1 = 868$ \boxed{(f \circ g)(x) = 48x^2 - 72x + 28, \quad (g \circ f)(x) = 12x^2 + 1, \quad (f \circ g)(5) = 868}$ --- ### Problem 5 $f(x) = 2x^2 - x + 5$, $g(x) = x + 4$ 1. $(f \circ g)(x) = f(g(x)) = f(x + 4) = 2(x + 4)^2 - (x + 4) + 5$ Expand: $$(x + 4)^2 = x^2 + 8x + 16$$ So, $$(f \circ g)(x) = 2(x^2 + 8x + 16) - x - 4 + 5 = 2x^2 + 16x + 32 - x + 1 = 2x^2 + 15x + 33$$ 2. $(g \circ f)(x) = g(f(x)) = g(2x^2 - x + 5) = (2x^2 - x + 5) + 4 = 2x^2 - x + 9$ 3. $(f \circ g)(5) = f(g(5)) = f(5 + 4) = f(9) = 2(9)^2 - 9 + 5 = 2(81) - 9 + 5 = 162 - 9 + 5 = 158$ \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 $f(x) = 3x$, $g(x) = x + 4$, $h(x) = x^2 - 1$ Calculate stepwise: 1. $h(5) = 5^2 - 1 = 25 - 1 = 24$ 2. $g(h(5)) = g(24) = 24 + 4 = 28$ 3. $f(g(h(5))) = f(28) = 3 \times 28 = 84$ \boxed{(f \circ g \circ h)(5) = 84}$ --- 2. Operations on Functions: Given $f$ and $g$, find indicated operations and simplify. --- ### Problem 1 $f(x) = 3x^2 - 6x + 14$, $g(x) = x - 4$ 1. $(f + g)(x) = f(x) + g(x) = (3x^2 - 6x + 14) + (x - 4) = 3x^2 - 5x + 10$ 2. $(f - g)(x) = f(x) - g(x) = (3x^2 - 6x + 14) - (x - 4) = 3x^2 - 7x + 18$ 3. $(f + g)(4) = f(4) + g(4)$ Calculate: $f(4) = 3(4)^2 - 6(4) + 14 = 48 - 24 + 14 = 38$ $g(4) = 4 - 4 = 0$ So, $(f + g)(4) = 38 + 0 = 38$ \boxed{(f + g)(x) = 3x^2 - 5x + 10, \quad (f - g)(x) = 3x^2 - 7x + 18, \quad (f + g)(4) = 38}$ --- ### Problem 2 $f(x) = \frac{3x + 1}{x}$, $g(x) = \frac{3}{x}$ 1. $(f + g)(x) = \frac{3x + 1}{x} + \frac{3}{x} = \frac{3x + 1 + 3}{x} = \frac{3x + 4}{x}$ 2. $(f - g)(x) = \frac{3x + 1}{x} - \frac{3}{x} = \frac{3x + 1 - 3}{x} = \frac{3x - 2}{x}$ 3. $(f / g)(x) = \frac{f(x)}{g(x)} = \frac{\frac{3x + 1}{x}}{\frac{3}{x}} = \frac{3x + 1}{x} \times \frac{x}{3} = \frac{3x + 1}{3}$ \boxed{(f + g)(x) = \frac{3x + 4}{x}, \quad (f - g)(x) = \frac{3x - 2}{x}, \quad (f / g)(x) = \frac{3x + 1}{3}}$ --- ### Problem 3 $f(x) = \frac{x}{4}$, $g(x) = \frac{x}{4}$ 1. $(f + g)(x) = \frac{x}{4} + \frac{x}{4} = \frac{2x}{4} = \frac{x}{2}$ 2. $(f - g)(x) = \frac{x}{4} - \frac{x}{4} = 0$ 3. $(f / g)(x) = \frac{\frac{x}{4}}{\frac{x}{4}} = 1$ (for $x \neq 0$) \boxed{(f + g)(x) = \frac{x}{2}, \quad (f - g)(x) = 0, \quad (f / g)(x) = 1}$ --- ### Problem 4 $f(x) = 2x + 1$, $g(x) = 4x - 3$ 1. $(f + g)(x) = (2x + 1) + (4x - 3) = 6x - 2$ 2. $(f - g)(x) = (2x + 1) - (4x - 3) = 2x + 1 - 4x + 3 = -2x + 4$ 3. $(fg)(x) = f(x) \times g(x) = (2x + 1)(4x - 3) = 8x^2 - 6x + 4x - 3 = 8x^2 - 2x - 3$ \boxed{(f + g)(x) = 6x - 2, \quad (f - g)(x) = -2x + 4, \quad (fg)(x) = 8x^2 - 2x - 3}$ --- ### Problem 5 $f(x) = x^2 - x + 1$, $g(x) = x - 1$ 1. $(f + g)(x) = (x^2 - x + 1) + (x - 1) = x^2$ 2. $(f - g)(x) = (x^2 - x + 1) - (x - 1) = x^2 - x + 1 - x + 1 = x^2 - 2x + 2$ 3. $(f / g)(x) = \frac{x^2 - x + 1}{x - 1}$ This fraction cannot be simplified further. \boxed{(f + g)(x) = x^2, \quad (f - g)(x) = x^2 - 2x + 2, \quad (f / g)(x) = \frac{x^2 - x + 1}{x - 1}}$ --- ### Problem 6 $f(x) = 48x^6$, $g(x) = 6x^4$ 1. $(fg)(x) = f(x) \times g(x) = 48x^6 \times 6x^4 = 288x^{10}$ 2. $(f / g)(x) = \frac{48x^6}{6x^4} = 8x^{6-4} = 8x^2$ 3. $(f / g)(6) = 8(6)^2 = 8 \times 36 = 288$ \boxed{(fg)(x) = 288x^{10}, \quad (f / g)(x) = 8x^2, \quad (f / g)(6) = 288}$