Math

• Notes
• HW 2.6
• HW 2.7
• HW 3.1
• HW 3.2
• HW 3.3
• HW 3.4
• HW 3.5
• HW 4.1
• HW 4.2
• HW 4.3
• HW 4.4
• HW 4.5
• Exam November 9th
• Exam November 9th Corrections
• Test Review 03-08-2016

• Chem 7.31 - 7.45

Q1 Find each of the following on the graph shown

(a) On what intervals is the function increasing?

The function is increasing on $(-3,3)\Cup(5,7)$

(b) On what intervals is the function decreasing?

The function is decreasing on $(3,5)$

(c) On what intervals is the function constant?

The function is constant on $(-7,-4)$

(d) What is the domain of the function?
(http://patrickjmt.com/finding-domain-and-range-of-a-function-using-a-graph/)

The domain of the function is $[-7,-4],[-3,7]$

(e) What is the range of the function?
(http://patrickjmt.com/finding-domain-and-range-of-a-function-using-a-graph/)

The range of the function is $[-6,-5],[6,6]$

Q2 Solve the following system of equations:

$\space \space 3x-2y+ \space 5z=4$
$\space 12x+9y+54z=0$
$-3x+\space y \space \space -7z=1$

Multiply first line by $4$:

$\space \space 3x-2y+ \space 5z=4$
$12x-8y+20z = 16$

Subtract $12x-8y+20z = 16$ from second line:

$\space 12x+9y+54z=0$
$12x-8y+20z = 16$

$=17y+34z=-16$

$\color{red}{12x-8y+20z = 16\\ 17y+34z=-16 \\ -3x+\space y \space \space -7z=1}$

Multiply last line by 4:

$-3x+\space y \space \space -7z=1$

$-12x+\space 4y \space \space -28z=4$

$\color{red}{12x-8y+20z = 16\\ 17y+34z=-16 \\ -12x+\space 4y \space \space -28z=4}$

Add first and third line:

$12x-8y+20z = 16$
$-12x+\space 4y \space \space -28z=4$

$=-4y -8z = 20$

$\color{red}{12x-8y+20z = 16 \\ 17y+34z=-16 \\ -4y -8z = 20 }$

Multiply second line by 4 and third line by 17 (to get LCM 136):

$(17y+34z)4=(-16)4$
$= 68y + 136z=-64$

$(-4y -8z)17 = 20(17)$
$= -68y - 136z = 340$

Add the two equations:

$68y + 136z=-64$
$-68 - 136z = 340$

$0=276$

$0 \ne 276$, therefore the system has no solution.

Q3

$A+B+C=3200$
$A+B=1900$
$A+C=2100$

$A = A$
$B = 1900-A$
$C = 2100-A$
$A+B+C=3200$
$A +(1900-A) + (2100-A) = 3200$

$A-2A=3200-1900-2100$
$-A=-800$
$\color{blue}{A=800}$

$B=1900-800$
$\color{blue}{B=1100}$

$C=2100-800$
$\color{blue}{B=1300}$

Q4 Use the graph of $y=f(x)$ shown on each graph to graph the function $g(x)$

$g(x)=f( \frac{1}{2}x)$

$y=f(x \cdot c )$

$x=-3,y=-1$
$g(-3)=f(-3 \cdot 1/2)$

$g(x)=f(x)+2$

The y-coordinates of the graph of $y=f(x)$ are shifted upwards by 2 points.

$x=-3, y=-1$
$g(x) = y+2$
$g(-3)=-1+2$
$g(-3)=1$

Q5 The domain of each of the following functions. Write answers in interval notation.

$f(x)=\sqrt{3x+7}$

$3x+7\ge \:0$

$\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides}$
$3x+7-7\ge \:0-7$
$3x\ge \:-7$

$\mathrm{Divide\:both\:sides\:by\:}3$
$\frac{3x}{3}\ge \frac{-7}{3}$

$x\ge \:-\frac{7}{3}$

$\color{blue}{\:[-\frac{7}{3},\:\infty \:)}$

$f(x)=\frac{9}{x^2-5x+6}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{9}{x^2-5x+6}\mathrm{\:and\:compare\:to\:zero}$

$x^2-5x+6=0$

$\mathrm{Factor}\:x^2-5x+6$

$x^2-5x+6$

$=\left(x^2-2x\right)+\left(-3x+6\right)$

$\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-2x\mathrm{:\quad }x\left(x-2\right)$
$\mathrm{Factor\:out\:}-3\mathrm{\:from\:}-3x+6\mathrm{:\quad }-3\left(x-2\right)$

$=x\left(x-2\right)-3\left(x-2\right)$

$\mathrm{Factor\:out\:common\:term\:}\left(x-2\right)$
$=\left(x-2\right)\left(x-3\right)$

$\left(x-2\right)\left(x-3\right)=0$

$x-2=0$
$x=2$

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

$x=3,\:x=2$

$\color{blue}{\left(-\infty \:,\:2\right)\cup \left(2,\:3\right)\cup \left(3,\:\infty \:\right)}$

Q8 Given $f(x)=3x^2-2x+4$ and $g(x)=x^2$, find each combination of these functions, simplify and state the domain.

$(f+g)(x)$

$(f-g)(x)$

$(fg)(x)$

$\mathrm{Assign\:the\:actual\:function}\:f=3x^2-2x+4$
$=\left(\left(3x^2-2x+4\right)g\left(x\right)\right)\left(x\right)$

$\mathrm{Assign\:the\:actual\:function}\:g=x^2$
$=\left(\left(3x^2-2x+4\right)x^2\right)\left(x\right)$
$$

$(f/g)(x)$

$\mathrm{Assign\:the\:actual\:function}\:g=x^2$
$=\left(\frac{f\left(x\right)}{x^2}\right)\left(x\right)$

$\mathrm{Assign\:the\:actual\:function}\:f=3x^2-2x+4$
$=\left(\frac{3x^2-2x+4}{x^2}\right)\left(x\right)$

$\color{blue}{=\frac{3x^2-2x+4}{x^2}}$

$(f \circ g)(x)$

$\mathrm{For}\:f=3x^2-2x+4\:\mathrm{substitute}\:x\:\mathrm{with}\:x^2$
$=3\left(x^2\right)^2-2x^2+4$

$\color{blue}{=3x^4-2x^2+4}$

Find $(f \circ g)(x)$

$f(x)=6x+11$
$g(x)=2x-1$

$\mathrm{For}\:f=6x+11\:\mathrm{substitute}\:x\:\mathrm{with}\:2x-1$
$=6\left(2x-1\right)+11$

$\mathrm{Expand}\:6\left(2x-1\right)+11$
$=6\cdot \:2x+6\left(-1\right)$
$=12x-6$
$=12x-6+11$
$\color{blue}{=12x+5}$

$f(x)=2x^2+3$
$g(x)=4x-1$

$f \circ g(x) =$

$\mathrm{For}\:f=2x^2+3\:\mathrm{substitute}\:x\:\mathrm{with}\:4x-1$
$=2\left(4x-1\right)^2+3$

Expand $2\left(4x-1\right)^2+3$

$=2\left(16x^2-8x+1\right)+3$

$=32x^2-16x+2+3$

$\color{blue}{=32x^2-16x+5}$

$g \circ f (x) =$

$\mathrm{For}\:g=4x-1\:\mathrm{substitute}\:x\:\mathrm{with}\:2x^2+3$

$=4\left(2x^2+3\right)-1$

Expand $4\left(2x^2+3\right)-1$

$=8x^2+12-1$

$\color{blue}{=8x^2+11}$

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