### Present Remotely

Send the link below via email or IM

CopyPresent to your audience

Start remote presentation- Invited audience members
**will follow you**as you navigate and present - People invited to a presentation
**do not need a Prezi account** - This link expires
**10 minutes**after you close the presentation - A maximum of
**30 users**can follow your presentation - Learn more about this feature in our knowledge base article

# 8.4 Inequalities of Combined Functions

No description

by

Tweet## Jenny Wong

on 8 January 2013#### Transcript of 8.4 Inequalities of Combined Functions

By: Cheng, Mina, Terrence, Jenny, Ocean 8.4: Inequalities of Combined Functions For inequalities of combined functions, you will be given 2 functions and will be asked to solve for a specific region.

For example: solve for when f(x) > g(x).

There are three ways to solve these inequalities:

1. Compare the Functions Visually

2. Analyze the Difference Function

3. Analyze the Quotient Function Intro Example One Example Two Which method is best? Example Three Let f(x) = x² and g(x) = 2x + 3.

Find the region in which

f(x) > g(x). a) Compare the Functions visually by graphing both on the same set of axis. b) Solve by Analyzing the Difference Function. Note: When solving by analyzing the difference function, you subtract one of the functions from both sides of the inequality.

f(x) > g(x)

f(x) - g(x) > g(x) - g(x)

f(x) - g(x) > 0

f(x) = x² and g(x) = 2x + 3

x² - (2x + 3) > 0 c) Solve by Analyzing the Quotient Function. Note: When solving by analyzing the Quotient function, you divide both sides of the inequality by one of the two functions.

Start by graphing the quotient function:

f(x) = x² and g(x) = 2x + 3

x²/(2x+3) > 1 Hint: It could be useful to graph y=1 to help visualize the region or you could subtract one from the entire function and use the x axis instead. When analyzing a quotient function, if the function that is dividing the other is negative, the inequality sign switches around. This means that you need to look out for the following two cases (in this case, the function you are dividing is g(x)):

If g(x) is negative, you are required to look for regions in which f(x)/(g(x)) < 1 (inequality sign switches)

If g(x) is positive, you are looking for regions in which f(x)/(g(x)) > 1(inequality sign does not switch) f(x) > g(x)

f(x)/(g(x)) > g(x)/(g(x))

Since g(x) = 2x + 3

Therefore, when x > -3/2, you are looking for regions in which

And when x < -3/2, you are looking for regions in which

Therefore, (-3/2, -1) U (3, ∞) U (-∞, -3/2).

(-∞, -1) U (3, ∞)

Check if f(x) > g(x) at VA

f(-3/2) = 9/4

g(-3/2) = 0

Therefore, f(x) > g(x) when x = -3/2 (-∞,-1) U (3, ∞) (-∞,-1) U (3, ∞) HA: y = 1 OA: y = (½)x - 3/4 VA: x = - 3/2 f(x) g(x) By: Cheng, Mina, Terrence, Jenny, Ocean a) Given f(x) = x² - 4 and

g(x) = x² - 5x + 4, solve for f(x) > g(x) by analyzing the difference function. (8/5, 0) (8/5, ∞) b) Given f(x) = x² - 4 and

g(x) = x² - 5x + 4, solve for f(x) > g(x) by analyzing the quotient function. VA: x = 4 VA: x = 1 HA: y = 1 (8/5, 1) g(x) > 0 when x > 4, when x < 1

g(x) < 0 when 1 < x < 4

Therefore, (4, ∞) U (8/5, 4)

Check if f(x) > g(x) at VA: x = 4

f(4) = 12

g(4) = 0

f(4) > g(4)

Therefore, (8/5, ∞). i) f(x) = x² + 6x + 4

g(x) = x² - 10 ii) f(x) = (x + 1)^6

g(x) = (x + 1)^4 iii) f(x) = 1/x

g(x) = log x vi) f(x) = 2x² + 5x + 7

g(x) = x² + 3x + 6 v) f(x) = (x - 1)(x + 2)

g(x) = (x + 3)(x - 5) iv) f(x) = x^5

g(x) = x³ (-1, 1) (3, 1) (0, 0) (3, 9) (3, 0) (-1, 0) (0, -3) f(x)/(g(x)) < 1 f(x)/(g(x)) > 1 (0, 0) (1, -4) (0, -8) (2, 0) (-2, 0) (0, -1)

Full transcriptFor example: solve for when f(x) > g(x).

There are three ways to solve these inequalities:

1. Compare the Functions Visually

2. Analyze the Difference Function

3. Analyze the Quotient Function Intro Example One Example Two Which method is best? Example Three Let f(x) = x² and g(x) = 2x + 3.

Find the region in which

f(x) > g(x). a) Compare the Functions visually by graphing both on the same set of axis. b) Solve by Analyzing the Difference Function. Note: When solving by analyzing the difference function, you subtract one of the functions from both sides of the inequality.

f(x) > g(x)

f(x) - g(x) > g(x) - g(x)

f(x) - g(x) > 0

f(x) = x² and g(x) = 2x + 3

x² - (2x + 3) > 0 c) Solve by Analyzing the Quotient Function. Note: When solving by analyzing the Quotient function, you divide both sides of the inequality by one of the two functions.

Start by graphing the quotient function:

f(x) = x² and g(x) = 2x + 3

x²/(2x+3) > 1 Hint: It could be useful to graph y=1 to help visualize the region or you could subtract one from the entire function and use the x axis instead. When analyzing a quotient function, if the function that is dividing the other is negative, the inequality sign switches around. This means that you need to look out for the following two cases (in this case, the function you are dividing is g(x)):

If g(x) is negative, you are required to look for regions in which f(x)/(g(x)) < 1 (inequality sign switches)

If g(x) is positive, you are looking for regions in which f(x)/(g(x)) > 1(inequality sign does not switch) f(x) > g(x)

f(x)/(g(x)) > g(x)/(g(x))

Since g(x) = 2x + 3

Therefore, when x > -3/2, you are looking for regions in which

And when x < -3/2, you are looking for regions in which

Therefore, (-3/2, -1) U (3, ∞) U (-∞, -3/2).

(-∞, -1) U (3, ∞)

Check if f(x) > g(x) at VA

f(-3/2) = 9/4

g(-3/2) = 0

Therefore, f(x) > g(x) when x = -3/2 (-∞,-1) U (3, ∞) (-∞,-1) U (3, ∞) HA: y = 1 OA: y = (½)x - 3/4 VA: x = - 3/2 f(x) g(x) By: Cheng, Mina, Terrence, Jenny, Ocean a) Given f(x) = x² - 4 and

g(x) = x² - 5x + 4, solve for f(x) > g(x) by analyzing the difference function. (8/5, 0) (8/5, ∞) b) Given f(x) = x² - 4 and

g(x) = x² - 5x + 4, solve for f(x) > g(x) by analyzing the quotient function. VA: x = 4 VA: x = 1 HA: y = 1 (8/5, 1) g(x) > 0 when x > 4, when x < 1

g(x) < 0 when 1 < x < 4

Therefore, (4, ∞) U (8/5, 4)

Check if f(x) > g(x) at VA: x = 4

f(4) = 12

g(4) = 0

f(4) > g(4)

Therefore, (8/5, ∞). i) f(x) = x² + 6x + 4

g(x) = x² - 10 ii) f(x) = (x + 1)^6

g(x) = (x + 1)^4 iii) f(x) = 1/x

g(x) = log x vi) f(x) = 2x² + 5x + 7

g(x) = x² + 3x + 6 v) f(x) = (x - 1)(x + 2)

g(x) = (x + 3)(x - 5) iv) f(x) = x^5

g(x) = x³ (-1, 1) (3, 1) (0, 0) (3, 9) (3, 0) (-1, 0) (0, -3) f(x)/(g(x)) < 1 f(x)/(g(x)) > 1 (0, 0) (1, -4) (0, -8) (2, 0) (-2, 0) (0, -1)