Section 3.1 Functions Relation – a set of ordered pair of points ( x, y ). Function – a Relation where every x coordinate in the Domain corresponds to exactly one y coordinate in the Range. Domain – a set of all x coordinates. Range – a set of all y coordinates. Dependent Variable – the variable that is isolated in an equation. Traditionally the y variable. Independent Variable – the variable that is not isolated in an equation. Traditionally the x variable.

Section 3.1 Functions “Y is a function of X.” – means that y is the Dep. Var. and x is the Indep. Var. Noted as y = f(x). Given that C is a function of t. What is the Dep. Var.? What is the Indep. Var.? How would you write the notation? C t Noted as C = f(t). Given that x is a function of y. What is the Dep. Var.? What is the Indep. Var.? How would you write the notation? x y Noted as x = f(y).

Section 3.1 Functions Determine if the correspondence is a Relation or a Function. Domain Range Domain Range Domain Range Relation x-coordinate repeated with different y-coordinate. Function Neither The y-coordinate of 9 is not matched with a x-coordinate.

Section 3.1 Functions Determine if the correspondence is a Relation or a Function. {( 2, 5 ), ( 3, 15 ), ( 4, -3 ), ( -2, 1 ), ( 1, 6 )} {( 1, 3 ), ( 2, 1 ), ( 3, 5 ), ( -2, 7 ), ( 1, -3 )} How to write equations as a function? Relation Function Solve for y! – 2x= – 2x No x – coordinates repeated with a different y – coordinate. Repeating x – coordinates with different y – coordinates. Implicit form Explicit form x = – 5, y = 1 – 2(– 5) y = = 11 y = – 7, – 7 = 1 – 2x – 1 = – 1 – 8 = – 2x – 2 x = 4 1. y can not be raised to an even power! 2. y can not be in absolute value bars! Find the value of y if x = -5. Find the value of x if y = -7

= 2(3) 2 + 3(3) For the function f denoted by, evaluate = 2(9) + 9 Substitute 3 in for x and always use ( )’s Simplify by the right side by the Order of Operations Rules. a.a. – x 3 = – x 3 Replace y with g(x). y is to an odd power … solve for y. y is to an even power … NOT A FUNCTION! = This means that if the x-coordinate is 3, then the y-coordinate is 27 or the point (3, 27).

= [ 2(3) 2 + 3(3) ] + For the function f denoted by, evaluate = 27 + Each function notation must be placed in [ ]’s because functions are at the “P” in PEMDAS. Combine Like Terms…CLT b.b. = 3(2x 2 ) + 3(3x) Distribute the 3 into the [ ]’s c.c.

= 2(- x) 2 + 3(- x) For the function f denoted by, evaluate Multiply the function by -1. Square the –x and multiply 3 and –x. d.d. Notice that this changes the sign on every term. e.e. Substitute –x in for every x using ( )’s. Substitute 3x in for every x using ( )’s. = 2(3x) 2 + 3(3x) Multiply 2 and 9x 2. f.f. = 2(9x 2 ) + 9x Square the (3x) and multiply 3 and (3x). = -1 [ 2x 2 + 3x ]

= 2(x + 3) 2 + 3(x + 3) For the function f denoted by, evaluate Substitute x + 3 in for every x using ( )’s. The binomial being squared should be written out twice and FOILed. Distribute the 3 into (x + 3). g.g. F = 2(x + 3)(x + 3) + 3(x + 3) = 2(x 2 + 3x + 3x + 9) + 3x + 9 IOL = 2x 2 + 6x + 6x x + 9 Distribute the 2 and CLT.

= [ 2(x + h) 2 + 3(x + h) ] – [ 2x 2 + 3x ] For the function f denoted by, evaluate This formula is known as the DIFFERENCE QUOTIENT. Test ?! Simplify the top first. We will save space by leaving the bottom out. h.h. = 2(x + h)(x + h) + 3(x + h) – 2x 2 – 3x = 2(x 2 + 2hx + h 2 ) + 3x + 3h – 2x 2 – 3x = 2x 2 + 4hx + 2h 2 + 3x + 3h – 2x 2 – 3x Distribute the 2 and that all non-h-terms cancel. The binomial being squared should be written out twice and FOILed. Distribute the 3 into (x + h) the minus sign into [ ]’s Factor the h as the GCF.Cancel the h’s.

Find the DOMAIN of a function defined by an equation. No restrictions, D : All Real Nbrs. Ask the question, if you can have negative values, zero, or positive values in the function? a.a. Set the denominator not equal to zero to find the restrictions. c.c. b.b. STOP! We can’t have zero in the bottom. D : x = + 2 This is called an implied domain, which means All Real Nbrs. except for + 2. STOP! We can’t take the square root of a negative value. Set the radicand, the expression in the radical symbol, greater than or equal to zero to make sure our variable generates a positive value. D : t < 3 or (-, 3]

Find the DOMAIN of a function defined by an equation. Ask the question, if you can have negative values, zero, or positive values in the function? Set the radicand greater than or equal to zero, but remember the denominator can not be equal to zero as well. This will find the restrictions. e.e. d.d. STOP! We can’t have zero in the bottom and take the square root of a negative. D : x > 1 or (1, ) STOP! We can’t have zero in the bottom. D : (-, ) or All Real Nbrs. Set the denominator not equal to zero to find the restrictions. Not possible, x 2 is always a non-negative value.

Find the DOMAIN of a function defined by an equation. Ask the question, if you can have negative values, zero, or positive values in the function? f.f. STOP! We can’t have zero in the bottom. Set the denominator not equal to zero to find the restrictions. STOP! We can’t have negatives in the square root.

Find the domain of the function, which represents the area of a circle with radius r. Ask the question, if you can have negative values, zero, or positive values in the function? STOP! We can’t have zero or negatives for the Area or the radius. D : r > 0 or (0, ) The Algebra of functions. Sum, Difference, Product, and Quotient of Two Functions SUM :DIFFERENCE: PRODUCT : QUOTIENT :

The Algebra of functions. Sum, Difference, Product, and Quotient of Two Functions SUM :DIFFERENCE: PRODUCT : QUOTIENT : Let and. Determine the functions. a.a. b.b. = f(2) + g(2) Substitute 2 into both functions and simplify each one separately. Next, perform the operation on the two values. = [ (2) ] + [ 3(2) + 5 ] = [ ] + [ ] = = f(5) – g(5) = [ (5) ] – [ 3(5) + 5 ] = [ ] – [ ] = 34 – 20

The Algebra of functions. Sum, Difference, Product, and Quotient of Two Functions SUM :DIFFERENCE: PRODUCT : QUOTIENT : Let and. Determine the functions. c.c. d.d. = f(-2) * g(-2) Substitute 2 into both functions and simplify each one separately. Next, perform the operation on the two values. = [ (-2) ] * [ 3(-2) + 5 ] = [ ] * [ ] = 13 * (-1) f(1) [ (1) ] [ ] 10 g(1) = = == [ 3(1) + 5 ] [ ] 8

Let and. Determine the functions. e.e. f.f. = f(x) + g(x) Substitute in both functions and CLT. Beware of subtraction, distribute the minus sign. = [ x ] + [ 3x + 5 ] = g(x) – f(x) = [ 3x + 5 ] – [ x ] = 3x + 5 – x 2 – 9 g.g. h.h. = f(x) * g(x) = [ x ] * [ 3x + 5 ] f(x) f(x) [ x ] g(x)g(x) == [ 3x + 5 ] ; x = 0i.i. f(x) f(x) g(x)g(x) = [ 3x + 5 ] = [ x ] Not Possible to equal zero.

Let and. Determine the functions. a.a. b.b. = j(x) + k(x) = j(x) – k(x) (x – 1) Bring in both domain restrictions. Check for canceling. Warning! Find domain restrictions. D : x = -2D : x = 1 1 (x + 2) =+ x (x – 1) Find LCD and add. (x – 1) (x + 2) = x – 1 + x 2 + 2x (x + 2)(x – 1) (x – 1) Bring in both domain restrictions. Check for canceling. 1 (x + 2) =– x (x – 1) Find LCD and subtract. (x – 1) (x + 2) = x – 1 – x 2 – 2x (x + 2)(x – 1) Distribute minus sign. CLT on top.

Let and. Determine the functions. c.c. d.d. = j(x) * k(x) k(x) Bring in both domain restrictions. Check for canceling. D : x = -2D : x = 1 1 (x + 2) = * x (x – 1) Another look at the domain restrictions. 1 (x + 2) = x (x – 1) = Multiply tops and bottoms, leave denominator in factored form. New domain restriction! j(x) j(x) = = 1 (x + 2) x (x – 1) x 1 (x + 2) * 1 (x + 2) x (x – 1) -2 makes top denominator 0. 1 makes bottom denominator 0. 0 makes the denominator of BIG fraction 0.