# Welcome to the Webinar “Live” Review for Test 3 (MAC 1105)

## Presentation on theme: "Welcome to the Webinar “Live” Review for Test 3 (MAC 1105)"— Presentation transcript:

Welcome to the Webinar “Live” Review for Test 3 (MAC 1105)
We will start promptly at 8:00 pm Everyone will be placed on mute I am happy that you are able to join us 

“Live” Review for Test 3 (MAC 1105)
The following slides present sample problems similar to the test content, including the worked-out solutions. These are a few problems and do not include all the competencies tested on Test 3. To be prepared for the test, you must review the material from all the corresponding sections and follow the Study Guide posted on Blackboard!

The function whose graph is shown below is a one-to-one function.
True or False (-3, 3) (3, 3) Answer: False Does not pass the horizontal line test. In a 1-1 function different inputs will always result in different outputs. In our example, inputs -3 and 3 yield the same output, 3; The given function is not 1-1. Remember for a function, one input gives one output, and for a one-to-one function, one output gives one input.

The function f(x) = 2x + 5 is a one-to-one function. a. True b. False
y = 2x + 5 is a linear function Algebraically: Answer: True Passes horizontal line test; two different inputs will not yield the same output. Input -2 1/2 3 10 Output 1 5 6 11 25

Find the inverse of f(x) = {(-5, 6), (0, 9), (1/2, -6), (3, 4)}
We must check something first! What? Is this a one-to-one function? Your time to work!

Find the inverse of f(x) = {(-5, 6), (0, 9), (1/2, -6), (3, 4)}
This is a one-to-one function, since different inputs produce different outputs. Finding the inverse: Interchange the x-and-y coordinates. So, f -1(x) = {(6, -5), (9, 0), (-6, 1/2), (4, 3)}

Find the inverse function for: f(x) = x + 2
5 y = x + 2 Interchange variables: x = y + 2 Solve for new y: first multiply both sides by 5 (5)x = (y + 2) (5) 5x = y + 2   y = 5x So, f-1(x)= 5x - 2

Find the domain and range of the inverse:
{(-5, 6), (0, 9), (1/2, -6), (3, 4)} The domain of the inverse is the range of the original function.  The range of the inverse is the domain of the original function. So, Domain of inverse = 6; 9; -6; 4 Range of inverse = -5; 0; 1/2; 3

State whether the inverse exists: y = x2 – 5.
This is a parabola; it fails the horizontal line test, thus, it is not one-to-one. Attempting to find the “inverse” would result in y =  sqrt(x + 5) whose graph is where, with the exception of x = -5, every input would have two different outputs. Not a function! We would need to restrict the domain of the given function.

If f(x) = 2x – 1 and g(x) = x + 4, find (f – g)(x).
(f – g)(x) = f(x) – g(x) = (2x – 1) – (x + 4) = 2x – 1 – x – 4 = x – 5 Find (f  g)(x) (f  g)(x) = f(x)  g(x) = (2x – 1)(x + 4) = 2x2 + 7x – 4 Find (f/g)(x) (f/g)(x) = f(x)/g(x) = 2x – 1 x + 4 x  –4

If f(x) = 2x – 1 and g(x) = x + 4, find (f – g)(3).
(f – g)(x) = f(x) – g(x) Two approaches: Find (f – g)(x) and then find (f – g)(3). (f – g)(x) = (2x – 1) – (x + 4) = 2x – 1 – x – 4 = x – 5 So, (f – g)(3) = 3 – 5 = -2 OR Find f(3) and g(3) and then subtract. f(3) = 2(3) – 1 = and g(3) = = 7 So, (f – g)(3) = 5 – 7 = -2

Example: If f(x) = 2x + 3 and g(x) = x – 8
"g composed with f” That is, function f will be the input for function g. Example: If f(x) = 2x and g(x) = x – 8 then “2x + 3” will be the input for function g. Since g(x) = x – 8 g(f(x)) = (2x + 3) – 8 = 2x – 5

If f(x) = -x2 and g(x) = , find (f o g)(x)
(f o g)(x) = f(g(x)) That is, function g will be the input for function f. So, f(g(x)) = f( ) = - ( )2 = -(x + 2) = -x − 2

If f(x) = -5x and g(x) = x2 + 2, find (g o f)(3).
One approach: Calculate (g o f)(x) = g(f(x)) and then evaluate the resulting function when x = 3 g(f(x)) = g(-5x) = (-5x)2 + 2 = 25x2 + 2   Thus, (g o f)(3) = 25(3)2 + 2 = 227 Another approach: Since (g o f)(x) = g(f(x)), calculate f(3), then substitute the resulting value into g and evaluate: f(3) = -5(3) = -15 Then, g(-15) = (-15)2 + 2 = 227

Degree? Highest exponent is 3, therefore degree is 3 Leading term? -x³ Leading coefficient? -1 Constant term? -8

Use the graph to determine the following for the polynomial function:
Even or odd degree? Positive or negative leading coefficient? Number of turning points? Odd degree: One end of the graph points up; the other end points down. Negative: The right-hand end points down. Turning points: 2

Even or odd degree? Positive or negative leading coefficient?
Both ends point down: Even degree and negative leading coefficient

Determine the intercepts of the polynomial function
To find y-intercept: Let x = 0; find f(0). To find x-intercepts: Let y = 0; solve f(x) = 0 y-intercept: f(0) = (0 – 2)(30 – 1)(0 + 4) = (-2)(-1)(4) = 8 y = 8 or the ordered pair (0, 8) x-intercepts: (x – 2)(3x – 1)(x + 4) = 0 x – 2 = x – 1 = x + 4 = 0 x = 2, 1/3, -4 or the ordered pair (2, 0), (1/3, 0), (-4, 0)

The graph of f(x) = (x – 1)( x + 3)(x – 3)² will cross the x-axis at
True or False? The zeros of the function are: -3, 1, 3 1 with multiplicity 1 -3 with multiplicity 1 3 with multiplicity 2 If k is a zero of odd multiplicity, the graph will cross the x-axis at k. If k is a zero of even multiplicity, the graph will touch the x-axis and turn around at k. So, the answer is False, because x = 3 is a zero with an even multiplicity.

Determine the factored form of the equation of the polynomial
whose graph is shown below. x(x + 4)(x – 4) b. -x(x + 4)(x – 4) c. (x + 4)²(x – 4)² d. (x + 4)(x – 4)² First, one end points up, the other end points down, so it is an odd-degree polynomial. (Cannot be “c”) It crosses the origin, so x = 0 is a real zero. (Cannot be “d”) Right-hand end points down, so, the leading coefficient is negative. Answer: b

Find the domain of the rational function:
y =   x² – 49 The denominator cannot equal zero: x² – 49  0 (x + 7) (x – 7)  0 x  -7 and x  7 or in interval notation (-infinity -7) U (-7, 7) U (7, infinity)

Find the vertical asymptote(s) of
x = 2/5, 3 x = -2/5, -3 x = 11/5, -11/5 x = 0 5x² – 17x + 6 = 0 (5x – 2)(x – 3) = 0 5x – 2 = 0 or x – 3 = 0 x = 2/5, 3 Answer is: a y = x   x² – 17x + 6

Horizontal Asymptotes
Let m = degree of numerator and n = degree of denominator If m < n horizontal asymptote is line y = 0 (the x-axis) If m = n horizontal asymptote is ratio of leading coefficients If m > n no horizontal asymptote Find the horizontal asymptote, if it exists: a. y = x² b. y = x c. y = x   x² – 3x x² – 3x x² – 3x + 6 m = n So, y = 7 = 7 is a horizontal asymptote 1 m > n So, there is no horizontal asymptote c. m < n So, y = 0 is a horizontal asymptote

Find the domain and the x-intercepts:
Domain: x  0 or in interval notation (-infinity, 0) U (0, infinity) x-intercepts: Solve f(x) = 0, (x – 5) (x + 6) = 0 So, x-intercepts: 5, -6 or in ordered pair (5, 0),(-6, 0) NOTE: It is important that you follow the instructions on the test. If asked for ordered pair, do so! If you are not asked for ordered pair, then just write the values separated by commas (,) or semicolons (;) as specified on the question.

The average cost, A, per item of producing x number of a new product-
item is given by A(x) = x   x a. Find and interpret A(400). b. If the average cost per item were \$7.50, how many could they produce? c. Graph the average cost function using [0, 500, 50] by [0, 100, 10]. a. A(400) = (400) = 6.8   The average cost per item of producing 400 items is \$6.80. b x = 7.5   x (x) x = 7.5 (x)   x x = 7.5x x =  276 items

c. Graph the average cost function using [0, 500, 50] by [0, 100, 10].
y = x   x Y1 = ( x)/x Average cost Number of T-shirts

This "live" review session has covered content addressed on test 3, but you need to review chapters 5 and 7 and complete all the problems assigned on the study guide, so that you are well-prepared for the test.

Good luck  Do Well!

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