Math / Chem

Math

Chem

Chem 7.31 - 7.45

HW2.6 Q1

$f(x)=\frac{15}{x+8}$

$x +8 =0$
$x \ne -8$

$(-\infty, -8)\cup(-8,\infty)$

HW2.6 Q2

$f(x) = \frac{1}{x+14} + \frac{9}{x-4}$

$x \ne -14, x \ne 4$

$(-\infty, -14)\cup(-14,4)\cup(4,\infty)$

HW2.6 Q3

$f(x) = 3x-8, g(x) = x -1$

Find $f+g$ , $fg$ and $\frac{f}{g}$

$(f+g)(x) = 3x-8 + x-1 \\= \color{blue}{4x -9}$

$(f-g)(x) = (3x-8) - (x-1) \\= 3x -8 -x +1 \\= \color{blue}{2x -7}$

$(fg)(x) = (3x-8)(x-1) \\= 3x^2-3x-8x+8 \\= \color{blue}{3x^2 -11x +8}$

$(\frac{f}{g})(x) = \color{blue}{\frac{3x-8}{x-1}}$

$x-1=0$
$x = 1$

$(-\infty,1)\cup(1,\infty)$

HW2.6 Q4

$f(x) = 4x^2+25x+36$

$g(x) = x+4$

$(f+g)(x) = 4x^2+25x+36 + x +4 \\= \color{blue}{4x^2 +26x +40}$

$(f-g)(x) = (4x^2+25x+36) - (x+4)\\= 4x^2+25x+36-x-4 \\= \color{blue}{4x^2 +24x+32}$

$(fg)(x) = (4x^2+25x+36)(x+4)$

(x + 4)(4x²+ 25x + 36)

4x³ + 25x² +  36x
      16x² + 100x + 144
------------------------
4x³ + 41x² + 136x + 144

$= \color{blue}{4x^3 +41x^2+136x+144}$

$(\frac{f}{g})(x) = \color{black}{\frac{4x^2+25x+36}{x+4}}$

$4x^2+25x+36 \\= (4x+9)(x+4)$

$(\frac{f}{g})(x) = \color{black}{\frac{(4x+9)(x+4)}{x+4}}$

$= \color{blue}{4x+9}$

$x+4=0$
$x = -4$

$(-\infty,-4)\cup(-4,\infty)$

HW2.6 Q5

$f(x) = \frac{6x+5}{x^2-64}$

$g(x) = \frac{5x-3}{x^2-64}$

$(f+g)(x) = \frac{6x+5}{x^2-64} + \frac{5x-3}{x^2-64}$

$= \frac{6x+5 + 5x-3}{x^2-64}$

$= \color{blue}{\frac{11x+2}{x^2-64}}$

$x^2-64 = 0$

Solve by factoring:

$(x+8)(x-8)=0$
$x = 8, x = -8$

The domain of $(f+g)(x)$ is $(-\infty,-8)\cup(-8,8)\cup(8,\infty)$ .

$(f-g)(x) = (\frac{6x+5}{x^2-64}) - (\frac{5x-3}{x^2-64})$

$= \frac{(6x+5) - (5x-3)}{x^2-64}$

$= \frac{x+8}{x^2-64}$

Factor:

$= \frac{x+8}{(x+8)(x-8)}$

$= \color{blue}{\frac{1}{x-8}}$

$(fg)(x) = (\frac{6x+5}{x^2-64}) \times (\frac{5x-3}{x^2-64})$

$= \color{blue}{\frac{(6x+5)(5x-3)}{(x^2-64)^2}}$

The domain of $(fg)(x)$ is $(-\infty,-8)\cup(-8,8)\cup(8,\infty)$ .

$(f/g)(x) = (\frac{6x+5}{x^2-64}) \div(\frac{5x-3}{x^2-64})$

$= (\frac{6x+5}{x^2-64}) \times(\frac{x^2-64}{5x-3})$

$= \frac{(6x+5)(x^2-64)}{(5x-3)(x^2-64)}$

Cancel out the common factor $x^2-64$ :

$= \color{blue}{\frac{6x+5}{5x-3}}$

$5x-3=0$
$5x = 3$

$x=\frac{3}{5}$

$= \frac{(6x+5)(x+8)(x-8)}{(5x-3)(x+8)(x-8)}$

$x=-8, x=\frac{3}{5}, x=8$

The domain of $(f/g)(x)$ is $(-\infty,-8)\cup(-8,\frac{3}{5})\cup(\frac{3}{5},8)\cup(8,\infty)$ .

HW2.6 Q6

$f(x)=\sqrt{14-x}$

$g(x)=\sqrt{x-14}$

$(f+g)(x) =\sqrt{14-x}+\sqrt{x-14}$

$(f-g)(x) =\sqrt{14-x}-\sqrt{x-14}$

$(fg)(x) =\sqrt{14-x}\times\sqrt{x-14}$

$(f/g)(x) =\frac{\sqrt{14-x}}{\sqrt{x-14}}$

HW2.6 Q8

$f(x)=x^2+6$

$g(x)=x^2-8$

$(f \circ g)(x) = (x^2-8)^2+6$

$=(x^2-8)(x^2-8)+6\\=x^4-8x^2-8x^2+64+6\\=\color{blue}{x^4-16x^2+70}$

$(g \circ f)(x) = (x^2+6)^2-8$

$=(x^2+6)(x^2+6)-8\\=x^4+6x^2+6x^2+36-8\\=\color{blue}{x^4+12x^2+28}$

$(f \circ g)(4) = 4^4-16(4)^2+70$

$=256−256+70\\=\color{blue}{70}$

$(g \circ f)(4) = 4^4+12(4)^2+28$

$=256+192+28\\=\color{blue}{476}$

HW2.6 Q9

$f(x)=\sqrt{x}$

$g(x)=x-3$

$(f \circ g)(x)=\sqrt{x-3}$

$(g \circ f)(x)=\sqrt{x}-3$

$(f \circ g)(7)=\sqrt{7-3}\\=\sqrt{4}\\=2$

$(g \circ f)(7)=\sqrt{7}-3$

HW2.6 Q10

$f(x)=\frac{3}{x+6}$

$g(x)=\frac{5}{x}$

$(f \circ g)(x) = \frac{3x}{6x+5}$

$f(g(x))=\frac{3}{g(x)+6}$

$g(x) = -6$

$\frac{5}{x}=-6\\5=-6x\\\frac{5}{6}=-x\\x=-\frac{5}{6}$

Domain: $(-\infty,-\frac{5}{6})\cup(-\frac{5}{6},0)\cup(0,\infty)$

HW2.6 Q11