IT100 Session 2: - Functions

Simplification

Some useful relationships

\[\large\begin{array}{rcl} (x+a)(x+b) &=& x^2 + (a+b)x + ab\\ (x+a)(x+a) &=& x^2 + 2a + a^2\\ (x+a)(x-a) &=& x^2 - a^2\\ (ax+b)(cx+d)&=& ac x^2 + (ad+bc)x + bd\\ \end{array}\]

Simplifying using additive inverses

\[\large\begin{array}{rcl} x + 3 & = & 5 \\ x = x + 3 -(3) & = & 5 - (3) = 2 \\ & & \\ x + 7 & = & 2x + 3 \\ 7 = x + 7 - (x) & = & 2x + 3 - (x) = x + 3\\ 4 = 7 - (3) & = & x + 3 - (3) = x \\ \end{array}\]

Simplifying using reciprocal values

\[\large\begin{array}{rcl} 2 x & = & 16\\ {2x \over (2)} & = & {16 \over (2)}\\ x & = & 8 \\ \end{array}\]

\[\large\begin{array}{rcl} {3\over x} &= & 36\\ 1 = {3(x)\over (3)x} &= & {36(x) \over (3)} = 12x\\ {1\over 12} = {1\over (12)} &=& {12x\over (12)} = x\\ \end{array}\]

Case 1

\[\large\begin{array}{c} \left({10h^2 − 9h − 9\over 2h^2 − 19h + 24}\right)\left({h^2 − 16h + 64\over 5h^2 − 37h − 24}\right)\\ {(5h+3)(2h-3)(h-8)(h-8)\over (2h-3)(h-8)(5h+3)(h-8)}\\ {(5h+3)(2h-3)(h-8)(h-8)\over (5h+3)(2h-3)(h-8)(h-8)}\\ 1\\ \end{array}\]

Case 2

\[\large\begin{array}{c} \left({x^2 − x − 6\over x^2 + x − 6}\right) \left({x^2 + 8x + 15\over x^2 − 9}\right)\\ {(x-3)(x+2)(x+3)(x+5)\over (x+3)(x-2)(x+3)(x-3)}\\ {(x+2)(x+5)\over (x-2)(x+3)}= {x^2 +7x + 10\over x^2 + x - 6}\\ \end{array}\]

Challenge

Simplify the following:
\[\large\begin{array}{c} \left({y^2 + 10y + 25\over y^2 + 11y + 30}\right)\\ {(y+5)(y+5)\over (y+5)(y+6)}\\ {y+5\over y+6}\\ \end{array}\] \[\large\begin{array}{c} \left({2x^2 + 7x − 4\over 4x^2 + 2x − 2}\right)\\ {(2x-1)(x+4) \over (2x-1)(2x+2)}\\ {x+4 \over 2x+2} = {x+4 \over 2(x+1)}\\ \end{array}\]

\[\large\begin{array}{c} \left({c^2 + 2c − 24\over c^2 + 12c + 36}\right)\left({c^2 − 10c + 24\over c^2 − 8c + 16}\right)\\ {(c-4)(c+6)(c-4)(c-6) \over (c+6)(c+6)(c-4)(c-4)}\\ {(c-6) \over (c+6)}\\ \end{array}\]

Algebraic Strategies

Algebra

• from Arabic “al-jabr”
  meaning “reunion of broken parts”

• a systematic application of the basic rules to simplify and solve equations

Strategy for simultaneous equations

1. Remove common multiples and simplify the equations separately
2. Transform the equations placing variables on one side of the equals and constants on the other
3. Combine terms
4. Solve for one variable
5. Substitute the value of the known variable to find the values of the others

Case 1

\[\large\begin{array}{rcl} 3x - 2y = 57; & & x + y = 44\\ & & \\ 2x + 2y=(2)(x + y) &=& (2)\times 44 = 88\\ & & \\ 5x = (3x-2y)+ 2x +2y &= & 57 + 88 = 145\\ x=5x/5&=&145 /5 = 29\\ & & \\ 29 + y &=& 44\\ y =29 + y -(29) & = & 44- (29)= 15\\ \end{array}\]

Case 2

\[\large\begin{array}{rcl} 8x - 3y = 8; & & x + 2y = 20\\ & & \\ 8x +16y = 8(x + 2y) &=& 8\times 20=160\\ 19y = 8x +16y -(8x - 3y) &=& 160 - 8 = 152\\ y = 19y/19 & = & 155/19 = 8\\ & & \\ x + 16 =x +2 \times 8 &= & 20\\ x = x + 16 - (16) &=& 20 -(16) = 4\\ \end{array}\]

Challenge

\[\large\begin{array}{rcl} 3x + 7 & = & 5x - 3\\ 7 = 3x + 7 - (3x) & = & 5x - 3 - (3x)= 2x -3\\ 7 & = & 2x - 3\\ 10 = 7 + (3) & = & 2x - 3 + (3) = 2x \\ 5 = 10 / (2) & = & 2x /(2) = x \\ 5 &=& x\\ \end{array}\]

\[\large\begin{array}{rcl} {-2 \over (3x+5)} & = & 6x - 4\\ -2 = {-2\times (3x+5) \over (3x + 5)} & = & (6x - 4)(3x+5) = 18x^2 +18x-20\\ -2 & = & 18x^2 - 18x - 20\\ -1= -2/2 & = & (18x^2 - 18x - 20) / 2 = 9x^2 -9x -10\\ 0 = -1 + (1) &=& 9x^2 - 9x -10 + (1) \\ 0 = 0/(9) & = & 9x^2 -9x -9 /(9) = x^2 -x - 1\\ \end{array}\]

\[\large\begin{array}{rcl} x & = & {-b \pm \sqrt{b^2 - 4ac}\over 2a }\\ x & = & {-(-1) \pm \sqrt{-1^2 - 4(1)(-1)}\over 2}\\ x & = & {1 \pm \sqrt{1+4}\over 2} = {1 \pm \sqrt{5}\over 2}\\ \end{array}\]

\[\large\begin{array}{rcl} {2x + 7\over x - 2} &=& 27\\ 2x + 7 = {(2x + 7)(x-2)\over x - 2} &=& 27(x-2) = 27x - 54\\ 7 = 2x + 7 - (2x) & = &27x -54 - (2x)= 25x -54\\ 7 + (54) & = & 25x -54 + (54)\\ {61\over 25} = {61\over (25)} & = & {25x\over (25)} = x \\ \end{array}\]

\[\large\begin{array}{rcl} 2x+5& = & 6y -15\\ x+8& = & y + 6\\ & & \\ 2x+16 =2(x+8) &=& 2(y+6) = 2y +12\\ 2x+4 = 2x+16 -(12) &=& 2y +12 -(12) = 2y\\ 1= 2x +5 -(2x+4) &=& 6y -15 - 2y = 4y -15\\ 16 = 1 +(15) &=& 4y -15 + (15) = 4y\\ 4 = 16/(4) &=& 4y/(4) = y\\ & & \\ x+8&=& 4 + 6= 10\\ x = x+8 - (8) = 10 - (8) = 2\\ \end{array}\]

Functions vs Non-Functions

Function Definition

A function is a relation in which each possible input value leads to exactly one output value.

\[\large\begin{eqnarray} Input & \rightarrow & \hbox{Function} & \rightarrow & Output\\ x\qquad & & f(x) & & {}\qquad y\\ \hbox{domain} & & y = {x \over 5} & & \hbox{range}\\ \large [0, 1, 2 ... n] & & & & \large [0, 0.2, 0.4 ... n/5] \\ \end{eqnarray}\]

Common non-function relationships

Circle
\(\large 25= x^2 + y^2\)

Ellipse
\(\large 25 = 2x^2+ 5y^2\)

Hyperbola

\(\large 25= x^2 y^2\)

\(\large 25=\left(x+1\right) \left(x-5\right) \left(y-1\right) \left(y+4\right)\)

Egg and Heart Shapes

\[\large 25 = \frac{x^2}{.5} + \frac{5y^2}{1-0.1x}\]

\(\large\left(x^2+y^2-2ax)^2 = 4a^2(x^2+y^2)\)

Flowers

\[\large\left(x^2+y^2\right)^3 = 4x^2y^2\]

Challenge

• Determine why these are considered non-functions in this form?

• There is more than one y-value per value of x
• There are values of x that have no corresponding values of y

• Some values of y are indeterminable and cannot be caluculated

• Is there a domain or mathematical transform that would make any of these a function?

• Using absolute value helps some
• Limiting the domain to a specific set of values also helps sometimes

• No magic bullet!

• Answer the questions on pg 20: Numbers 52-69

• Give answers for Applied Exercises on pg 21: Numbers 88-90

Inverse Functions

An inverse function will convert output back to the corresponding input value.

\[\large\begin{eqnarray} Input & \rightarrow & \hbox{Function} & \rightarrow & Output\\ x\qquad & & f(x) & & {}\qquad y\\ & & & & \\ Output & \rightarrow & \hbox{Inverse Function} & \rightarrow & Input\\ y\qquad & & f\prime(y) & & {}\qquad x\\ \end{eqnarray}\]

Challenge

Input \(\large(x)\)	Function	Output \(\large (y)\)	Inverse Function	Reverted \(\large (x_2)\)	Does \(\large x = x_2\) ?
\(\Large [-5,-2,0,2,5]\)	\(\large y = 3x\)	\(\Large [-15,-6,0,6,15]\)	\(\large x_2 = y / 3\)	\(\Large [-5,-2,0,2,5]\)	True
\(\Large [-5,-2,0,2,5]\)	\(\large y = x^2\)	\(\Large [25,4,0,4,25]\)	\(\large x_2 = \sqrt{y}\)
\(\Large [-5,-2,0,2,5]\)	\(\large y = 3x + 4\)	\(\Large [-11,-2,4,10,19]\)
\(\Large [-5,-2,0,2,5]\)	\(\large y = 1/x\)
\(\Large [-5,-2,0,2,5]\)	$y =	x	/x$

Function Transforms

Function transforms

Transform	Formula	Parabola	Line	Normal Curve
(N) None	\(\large y = f(x)\)	\(\large y = x^2\)	\(\large y = 2x\)	\(\large y = {1\over \sigma\sqrt{2\pi}} e^{-{(x-\mu)^2\over 2 \sigma^2)}}\)
(A) Horizontal	\(\large y = f(x+A)\)	\(\large y = (x+A)^2\)	\(\large y = 2(x + A)\)	\(\large y = {1\over \sigma\sqrt{2\pi}} e^{-{(x+A-\mu)^2\over 2 \sigma^2)}}\)
(B) Vertical	\(\large y = f(x) + B\)	\(\large y = x^2 + B\)	\(\large y = 2x + B\)	\(\large y = B+ {1\over \sigma\sqrt{2\pi}} e^{-{(x-\mu)^2\over 2 \sigma^2)}}\)
(C) Amplitude	\(\large y = C\times f(x)\)	\(\large y = C x^2\)	\(\large y = C(2x)\)	\(\large y = {C\over \sigma\sqrt{2\pi}} e^{-{(x-\mu)^2\over 2 \sigma^2)}}\)

Graph of Transformations

transforms

transforms

Normal Graph

transforms

Challenge

• Compare the graph of \(\large f(x+A)\) to that of \(\large f(x)\)

• Compare the graph of \(\large f(x) + B\) to that of \(\large f(x)\)

• Compare the graph of \(\large f(x)\times C\) to that of \(\large f(x)\)

Topic for Unit 5

Linear functions Read Chapter 2 Precalculus