2. Functions and Function Notation
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Chapter 3
2.
Functions and Function Notation
Functions and their notations play a pivotal role in algebra. This lesson delves into the nuances of how functions can be combined, not just through basic operations like addition or subtraction, but also through compositions. Compositions of functions involve evaluating one function using another, offering a layered approach to understanding mathematical relationships. Function notation, on the other hand, provides a standardized way to represent these relationships. It's essential to grasp these concepts as they form the foundation for more advanced mathematical studies. By mastering function notation and understanding compositions, one can tackle complex problems with ease and confidence.
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| | 10 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Functions and Function Notation
Slide of 10
Once a relation is known to be a function, it is handy to write it in a way that directly gives essential information such as the independent and dependent variables. This is where function notation comes to the rescue. This lesson will explore some applications of function notation and how this way of representing functions helps combine different functions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Evaluating Functions at Another Function
As with algebraic expressions, functions can also be evaluated at a specific input. In the following applet, drag a function into each empty box to explore the outputs obtained when a function is evaluated at another function.
It can be seen that f(g(x))=g(f(x)), which suggests that the order does not affect the result. Is it always true? Is it true for f(h(x)) and h(f(x))?
Discussion
Representing Functions Using Function Notation
There are different ways to represent a function — using tables of values, mapping diagrams, graphs, and equations. However, aside from these, one of the most common ways is using function notation.
Concept
Function Notation
Function notation is a special way to write functions that explicitly shows that y is a function of x — in other words, that y depends on x. Function notation is symbolically expressed as y=f(x) and read 
y equals f of x.Equations that are functions can be written using function notation.
Equation y=-5x+4 Function Notation f(x)=-5x+4
Notice that y has been replaced by f(x). In function notation, x represents an element of the domain and f(x) represents the element of the range that corresponds to x. When written in function notation, the expression that describes how to convert an input into an output — the right-hand side expression — is called the function rule. Besides f, other letters such as g or h can be used to name the function. Similarly, letters other than x can name the independent variable.
Extra
Interpreting Function NotationTo interpret an equation given in function notation, it is necessary to understand what both sides of f(x)=k mean. For example, consider the following equation.
f(3)=12
Here, f(3) denotes that the function's input is x=3 and that 12 is the output corresponding to this input.
f(3)=12⇓The output of f when x=3 is 12.
Now, consider a different scenario. Let w(t)=200t be a function that describes the number of words Kevin reads in t minutes. The following statements are true for this function.
w(4)andw(t)=900
Here, w(4) is the number of words that Kevin reads in 4 minutes and can be found by evaluating the function for t=4. However, the input is not a particular number in the second statement. In such cases, the statement can be interpreted as a question.
w(t)=900⇓For which value of tis theoutput equal to 900?
Based on the context, the second statement asks how many minutes it takes Kevin to read 900 words. To find this value of t, the equation w(t)=900 has to be solved for t.
Discussion
Evaluating Functions and Finding a Certain Input
In the same way that an expression can be evaluated at a particular x-value, functions can also be evaluated at a specific input. Furthermore, it is also possible to determine the input that produces a specific output.
Method
Evaluating a Function
Evaluating a function involves determining the value of the function when its independent variable is set to a specific value. This is done by substituting the given input value for the variable and evaluating the function rule. As an example, consider the value of the following function when x=4.
f(x)=3x+4
To evaluate a function for a particular input, there are two steps to follow.
1
Substitute the Input for x
expand_more Substitute the given input value for every instance of the independent variable in the function. In this case, 4 is substituted for x.
2
Evaluate the Expression
expand_more Next, evaluate the resulting expression by performing the required operations. The simplified value will be the output for the given input value.
As shown, when the input is 4, the output of the function is 16.
Method
Finding the Input of a Function
Given a function, it is possible to find the input that produces a certain output. This is done by substituting the given output value for the dependent variable and then solving for the independent variable. For the following function, try finding the x-value for which f(x)=21.
f(x)=4x−3
To find the input that produces a certain output, there are two steps to follow.
1
Substitute the Output for f(x)
expand_more Start by substituting the given output value for f(x). In this case, it means substituting 21 for f(x).
2
Solve for x
expand_more Example
Modeling With Functions
For the last few days of the long holiday, Izabella and Emily visited a city they did not know. Before returning home, they went into a souvenir shop where all the souvenirs were priced the same. After picking out some gifts and just before paying, Emily found a $3 gift card in her purse.
The function C(s)=4.5s−3 represents the cost, in dollars, of buying s souvenirs with a $3 gift card.
a How much will the girls pay if they buy ten souvenirs?
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b How many souvenirs can the girls buy with $69?
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Hint
b Find the value of s for which the output is 69. Substitute 69 for C(s) and solve the resulting equation for s.
Solution
a Based on the given information, finding how much will the girls pay for buying ten souvenirs is equivalent to finding C(10). This can be done by substituting 10 for s into the function rule and simplifying the right-hand side.
b In this context, C(s) gives the cost of buying s souvenirs. Therefore, asking how many souvenirs the girls can buy with $69 is the same as asking for which value of s the output of C is equal to 69.
C(s)=69
To find such value of s, in the function rule, substitute 69 for C(s) and solve the resulting equation for s.
C(s)=4.5s−3
Substitute
C(s)=69
69=4.5s−3
▼
Solve for s
AddEqn
LHS+3=RHS+3
72=4.5s
DivEqn
LHS/4.5=RHS/4.5
4.572=s
CalcQuot
Calculate quotient
16=s
RearrangeEqn
Rearrange equation
s=16
Example
Evaluating Functions
While walking downtown, Emily and Izabella walked into a craft store and saw a beautiful wooden jewelry box. The craftsman said that the price of the box is half its volume, which is given by the function V(x)=9(x+1)−3(3−x), where x is the length of an inner side of the box — a measure that the girls have to provide.
a Emily wants to buy a box where the length of the inner side of the box is 1 inch. How much will she pay for this box?
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b Izabella paid $10.50 for her box. What is the length of the inner side of the box she asked for?
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Hint
a Start by finding the volume of the box for the measure Emily chose. To do this, find V(1). The price is half the volume.
b Use the given price to find the volume of Izabella's box. Then, substitute this volume for V(x) in the function rule and solve the resulting equation for x.
Solution
a Since the price depends on the volume, the volume of the box needs to be calculated first. Note that the volume of a box where the inner side is 1 inch long is given by V(1). Substitute 1 for x into the function rule and simplify the right-hand side.
V(x)=9(x+1)−3(3−x)
▼
Substitute 1 for x and simplify
Substitute
x=1
V(1)=9(1+1)−3(3−1)
AddSubTerms
Add and subtract terms
V(1)=9(2)−3(2)
Multiply
Multiply
V(1)=18−6
SubTerm
Subtract term
V(1)=12
b To find the length of the inner side of Izabella's box, the volume of the box needs to be calculated first. If the price is half the volume of the box, then the volume of the box is twice the price paid. Therefore, the volume of the box will be found using the fact that Izabella paid $10.50 for the box.
2⋅10.50=21in3
The volume of Izabella's box is 21 cubic inches. The next step is to find the value of x that produced this volume. To do so, substitute V(x) for 21 in the function rule and solve the equation for x.
V(x)=9(x+1)−3(3−x)
Substitute
V(x)=21
21=9(x+1)−3(3−x)
▼
Solve for x
Distr
Distribute 9and-3
21=9x+9−9+3x
AddSubTerms
Add and subtract terms
21=12x
DivEqn
LHS/12=RHS/12
1221=x
CalcQuot
Calculate quotient
1.75=x
RearrangeEqn
Rearrange equation
x=1.75
Pop Quiz
Evaluating Different Functions
Discussion
Combining Functions
Two functions f(x) and g(x) can be combined into a sum, a difference, a product, or a quotient through the following formulas.
(f+g)(x)(f−g)(x)(f⋅g)(x)(gf)(x)=f(x)+g(x)=f(x)−g(x)=f(x)⋅g(x)=g(x)f(x),g(x)=0
However, apart from these four operations, two functions can also be combined into a composition. Concept
Composite Function
A composite function, or a composition of functions, combines two or more functions, which produces a new function. In a composition, the outputs produced by one function are the inputs of the other function. The composition of the functions f and g is denoted as f(g(x)) or (f∘g)(x).
Performing the composition of two functions is similar to evaluating one function into the other. For example, let f(x)=2x+1 and g(x)=x−5. To find f(g(x)), the variable x in f(x)=2x+1 must be substituted with g(x).
Note that f(g(x)) makes sense only when the outputs of g belong to the domain of f. Also, be aware that the composition of functions is not commutative — that is, in general, f(g(x))=g(f(x)). This can be checked with the same two functions.
Just as changing the order of the machines in a factory could alter the final product, changing the order in which the functions are applied could produce different outputs. For example, here f(g(2)) and g(f(2)) are different values.
f(x)=2x+1
▼
Substitute g(x) for x and evaluate
Substitute
x=g(x)
f(g(x))=2g(x)+1
Substitute
f(g(x))=h(x)
h(x)=2g(x)+1
Substitute
g(x)=x−5
h(x)=2(x−5)+1
Distr
Distribute 2
h(x)=2x−10+1
AddTerms
Add terms
h(x)=2x−9
Extra
Parts of a CompositionExplore the parts of a composition by moving the magnifying glass.
Example
Applying Discounts Through Compositions of Functions
Izabella and Emily are shopping for pants. Both girls have a $3 discount coupon for the store. After choosing a pair of pants each, they discovered that the pants are 25% off the marked price. The girls could not agree which order of discount would save them more money, so the cashier allowed them to apply the discounts in whichever order they wanted.
The function f(x)=x−3 represents the effect of applying only the coupon to the marked price, while g(x)=43x represents the effect of applying only the store's sale discount to the marked price. In both functions, x represents the marked price.
a Write a function I(x) that represents how much Izabella will pay if she uses the coupon first and then applies the store's sale discount.
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b Write a function E(x) that represents how much Emily will pay if she uses the store's discount first and then applies the coupon.
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c Which of them saved more money?
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Hint
a Applying the coupon first means to apply f to the marked price. Then, g must be applied to the resulting output. Therefore, I(x) is equal to g(f(x)). To find how much Izabella paid, evaluate I(31).
b This time, apply the store's discount first — that is, start by applying g to the marked price. Then, apply f to the resulting output. In other words, E(x) equals f(g(x)). To determine the marked price, equate E(x) to the amount paid by Izabella. Then, solve the resulting equation for x.
c Compare the total amount paid by each girl and the marked price of the chosen pair of pants.
Solution
a Since Izabella decided to apply the coupon first, the function f will be applied first to the marked price.
g(x)=43x
Substitute
x=f(x)
g(f(x))=43f(x)
Substitute
g(f(x))=I(x)
I(x)=43f(x)
Substitute
f(x)=x−3
I(x)=43(x−3)
I(x)=43(x−3)
▼
Substitute 31 for x and evaluate
Substitute
x=31
I(31)=43(31−3)
SubTerm
Subtract term
I(31)=43(28)
MoveRightFacToNum
ca⋅b=ca⋅b
I(31)=484
CalcQuot
Calculate quotient
I(31)=21
b Emily decided to apply the discounts in the opposite order — that is, she chose to apply the store's sale discount first. For her purchase, the function g will be applied to the marked price.
f(x)=x−3
Substitute
x=g(x)
f(g(x))=g(x)−3
Substitute
f(g(x))=E(x)
E(x)=g(x)−3
Substitute
g(x)=43x
E(x)=43x−3
E(x)=43x−3
Substitute
E(x)=21
21=43x−3
▼
Solve for x
AddEqn
LHS+3=RHS+3
24=43x
MultEqn
LHS⋅4=RHS⋅4
96=3x
DivEqn
LHS/3=RHS/3
32=x
RearrangeEqn
Rearrange equation
x=32
c To determine who saved more money, it would help to organize the results obtained in a table.
| Marked Price | Amount Paid | |
|---|---|---|
| Izabella's Pants | $31 | $21 |
| Emily's Pants | $32 | $21 |
E(31)=43(31)−3=20.25
In contrast, if Emily had applied the discounts as Izabella did, she would have paid $21.75. Regardless of who applied the discounts better, both girls got a big discount on their purchase and went home happy.
Closure
A Special Case of Compositions of Functions
As shown, the composition of functions is not commutative, implying that, in general, f(g(x))=g(f(x)). However, for those functions that do commute, there is a special case. Consider, for example, the following pair of functions.
Next, perform the composition in the opposite order, that is, g(f(x)).
As can be seen, both compositions resulted in the same function.
f(x)=3x−6andg(x)=31x+2
For these two functions, start by finding f(g(x)).
f(x)=3x−6
▼
Substitute g(x) for x and simplify
f(31x+2)=3(31x+2)−6
f(31x+2)=x
g(x)=31x+2
▼
Substitute f(x) for x and simplify
g(3x−6)=31(3x−6)+2
▼
Simplify right-hand side
Distr
Distribute 31
g(3x−6)=31⋅3x−31⋅6+2
a⋅a1=1
g(3x−6)=1⋅x−31⋅6+2
MultByOne
a⋅1=a
g(3x−6)=x−31⋅6+2
MoveRightFacToNum
ca⋅b=ca⋅b
g(3x−6)=x−36+2
CalcQuot
Calculate quotient
g(3x−6)=x−2+2
AddTerms
Add terms
g(3x−6)=x
f(g(x))=g(f(x))=x
The important thing here is not the fact that the functions commute but that the result of the composition is simply x, the original input. In such cases, the functions are said to be inverses of each other. For more information, read about the inverse of a function.
While checking the storage room of the cinema where he works, Diego found an old projector and decided to test it. The distance the projector should be placed from the wall depends on the desired image height.
According to the instructions, the distance can be calculated using function d(h)=3h, where d is the distance between the projector and the wall and h is the height of the projected image.
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We know that Diego wants to project an image 2 meters high, which means that h=2.
To determine the distance at which he should place the projector, we evaluate the function d at h=2.
This means that Diego should place the projector 6 meters away from the wall to get an image 2 meters high.
We are told that Diego placed the projector 7.2 meters away from the wall and we want to calculate the height of the projected image.
In other words, we need to find the value of h that produces a distance of 7.2 meters. This means that we are asked for what value of h the output d is equal to 7.2. d(h) = 7.2 To find the value of h, we will substitute 7.2 for d(h) in the function rule and solve the resulting equation for h.
We found that the output is 7.2 when h=2.4. In other words, if Diego places the projector 7.2 meters from the wall, the resulting projected image will be 2.4 meters high.
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Emily left went jogging this morning. The distance she travels, in meters, depends on her jogging time, in minutes, and can be calculated using the function d(t)=5(25t+6)−3(t+10).
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We know that Emily jogged for 25 minutes. Therefore, we need to evaluate the distance function at t=25 to determine how far from home she is at that moment.
After jogging for 25 minutes, Emily will be 3050 meters away from her house.
We are told that the stadium is 4026 meters from Emily's house and we want to calculate the amount of time time it would take her to get there. In other words, we need to find the value of t that produces a distance of 4026 meters in the function rule. In mathematical terms, we are asked for what value of t is the output d equal to 4026. d(t) = 4026 To find the value of t, we will substitute 4026 for d(t) into the function rule and solve the resulting equation for t.
We found that the output is 4026 when t=33. In other words, it would take Emily 33 minutes to jog from her house to the stadium.
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f(x)g(x)=2x−5=2x+5+1
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["x+1"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord\">4<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord\">4<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["1"]}}
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We need to find the composition f(g(x)), so let's start by writing the corresponding function rules. f(x) & = 2x-5 [0.15cm] g(x) & = x+5/2+1 In this case, the required composition is equivalent to evaluating the function f(x) at g(x), which is done by substituting g(x) for x. Thus, let's do it!
Next, on the right-hand side, we substitute the x+52+1 for g(x) and simplify. Note that we do not need to make the substitution on the left-hand side expression.
In this case, we need to find the composition g(f(x)). We will proceed as in the previous part, but this time, in g(x)= x+52+1, we substitute x with f(x).
Next, on the right-hand side, we change f(x) for its function rule, which is 2x-5. Again, we do not need to make this change on the left-hand side expression.
We need to evaluate the functions we found in the previous parts at x=4. f(g(x)) &= x + 2 [1ex] g(f(x)) &= x + 1 Let's start with f(g(x)).
In a similar fashion, let's evaluate g(f(x)) at x=4.
Finally, we subtract g(f(4)) from f(g(4)). f(g(4)) - g(f(4)) = 6 - 5 ⇓ f(g(4)) - g(f(4)) = 1
Note that finding f(g(4)) is the same as finding g(4) and then, substituting the resulting value into f(x).
Now we have to evaluate f(x) at 112.
The value of g(f(4)) can be found by following the same procedure. We start by finding f(4).
Next, we evaluate g(x) at 3.
We have obtained that f(g(4))=6 and g(f(4))=5. Therefore, f(g(4)) - g(f(4)) = 1.
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f(x)g(x)=6−7x=3x+4
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-21x+22"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">1<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">1<\/span><span class=\"mclose\">)<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["42"]}}
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We need to find the composition f(g(x)). In this case, this is equivalent to evaluating the function f(x) at g(x), so let's do it.
Next, on the right-hand side, we substitute 3x+4 for g(x) and simplify.
In this case, we need to find the composition g(f(x)). We will proceed as in the previous part, but this time, in g(x)=3x+4, we substitute x for f(x).
Next, on the right-hand side, we change f(x) for its function rule, which is 6-7x.
We need to evaluate the functions we found in the previous parts at x=-1. f(g(x)) &= -21x - 22 [1ex] g(f(x)) &= -21x + 22 Let's start with f(g(x)).
In a similar fashion, let's evaluate g(f(x)) at x=-1.
Finally, let's add f(g(-1)) and g(f(-1)). f(g(-1)) + g(f(-1)) = -1 + 43 ⇓ f(g(-1)) + g(f(-1)) = 42
Having computed the compositions f(g(x)) and g(f(x)) is really helpful in order to find f(g(-1))+g(f(-1)). However, it can still be done without using the functions found in Parts A and B. Note that finding f(g(-1)) is the same as finding g(-1) and then substituting the resulting value into f(x).
Now we have to evaluate f(x) at 1.
We got that f(g(-1))=-1. The value of g(f(-1)) can be found following the same procedure. We will start by finding f(-1).
Next, we evaluate g(x) at 13.
Consequently, g(f(-1))=43. Therefore, f(g(-1)) + g(f(-1))=42.
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