2.2 notation and algebra of functions

Notation and Algebra of Functions

http://www.lahc.edu/math/precalculus/math_260a.html

Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50for 5 pizzas (delivered).

Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”.

Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”.

Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range).

The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered),

The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).

The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).So this procedure is a function.

The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).So this procedure is a function. Its domain is D = {1, 2, 3, …}

The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).So this procedure is a function. Its domain is D = {1, 2, 3, …} and its range is R = {various $-cost}.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver. The domain D =the range R =

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R =

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}.b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder.

Notation and Algebra of FunctionsExample A. Is the following procedure a function?If it’s a function, describe its domain and its range.a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}.b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder.This is not a function because the license holder may have many cousins so there might be several outputs.

Notation and Algebra of FunctionsIn mathematics, we often use the letters f, g, h,.. as names of functions.

Notation and Algebra of FunctionsIn mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to yfrom the domain D to the range R, we write that f(x) = y,

Notation and Algebra of FunctionsIn mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to yfrom the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:

the domain D the range R

Suppose f is a function that assigns x to yfrom the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:

x y = f(x)

the domain D the range R

x y = f(x)

the domain D the range RIn applications, functions are named after their purposes.

x y = f(x)

the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name”.

x y = f(x)

the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”,

then Licen#-to-name(123456) =

x y = f(x)

then Licen#-to-name(123456) =

the input x

x y = f(x)

then Licen#-to-name(123456) = “Joe Blow”.

the input x the output y

x y = f(x)

then Licen#-to-name(123456) = “Joe Blow”.

the input x the output y123456 “Joe Blow”

D ={Lic–numbers} R={Names}

Licen#-to-name

Notation and Algebra of FunctionsThe domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.

There are many ways to describe functions.

The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.

We may define functions by written instructions as in the case for the “Licen#-to-name” function.

We may define functions with tables, for example:x y=f(x

)–1 4 2 3 5 –3 6 4 7 2

We may define functions with tables, for example:x y=f(x

)–1 4 2 3 5 –3 6 4 7 2

We search the table for outputs so f(2) = 3, f(6) = 4, etc..

x y=f(x)

–1 4 2 3 5 –3 6 4 7 2

The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}.

We may define functions by written instructions as in the case for the “Licen#-to-name” function.We may define functions with tables, for example:

x y=f(x)

–1 4 2 3 5 –3 6 4 7 2

Note that f(–1) = f(6) = 4,

x y=f(x)

–1 4 2 3 5 –3 6 4 7 2

Note that f(–1) = f(6) = 4, so that a function may assign multiple inputs to the same output.

Notation and Algebra of FunctionsFunctions may be given graphically as the ones here:

For instance, Nominal Price(1975) $0.50

(1975, $0.50)

Nominal–price is the price that’s posted at the gas stations.

Functions may be given graphically as the ones here:

Domain (Nominal Price) = {year 1918 2005}For instance, Nominal Price(1975) $0.50

(1975, $0.50)

1918 2005

Domain (Nominal Price) = {year 1918 2005}Range (Nominal Price) = {$0.20$2.51}

For instance, Nominal Price(1975) $0.50

(1975, $0.50)

1918 2005

(1975, $0.50)

(1975, $1.85)

Adjusted Price(1975) $1.85

Adjusted–price is the inflation adjusted price in 2007–dollar.

Domain (Adjusted Price) = {year 1918 2005}Adjusted Price(1975) $1.85

(1975, $0.50)

(1975, $1.85)

1918 2005

Domain (Adjusted Price) = {year 1918 2005}Range (Adjusted Price) = {$1.25$3.50}

Adjusted Price(1975) $1.85

(1975, $0.50)

(1975, $1.85)

1918 2005

Notation and Algebra of FunctionsLet f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f.

Notation and Algebra of FunctionsLet f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t containtwo or more points whose x–coordinates are the samei.e. having multiple points lining up vertically.

Notation and Algebra of FunctionsLet f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t containtwo or more points whose x–coordinates are the samei.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out.

yA. B. C. D.

Notation and Algebra of FunctionsLet f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t containtwo or more points whose x–coordinates are the samei.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out.Hence A and B below are graphs of functions

A and B are functions because for each input x there is one output y.

A. B. C. D.

(x, y)

Notation and Algebra of FunctionsLet f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t containtwo or more points whose x–coordinates are the samei.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out.Hence A and B below are graphs of functions but C and D are not graphs of functions.

(x, y2)

A and B are functions because for each input x there is one output y.

A. B. C. D.

C and D are not since there are multiple points lining up vertically.

(x, y)

x(x, y1)

(x, y3)

Notation and Algebra of FunctionsMost functions are given by mathematical formulas.

For example,

f(X) = X2 – 2X + 3 = y

For example,

f(X) = X2 – 2X + 3 = y

name of the function

For example,

f(X) = X2 – 2X + 3 = y

name of the function

input box

For example,

f(X) = X2 – 2X + 3 = y

name of actual formulathe function

input box

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

The input box holds the input for the formula.

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

The input box holds the input for the formula.Hence f (2) means to replace x by (2) in the formula, so f(2) =

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

The input box holds the input for the formula.Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

The input box holds the input for the formula.Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.

For example,

f(X) = X2 – 2X + 3 = y

The output

input box

The input box holds the input for the formula.Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y. Real functions are functions whose domain and range are subsets of real numbers (no complex numbers).

Domain of Functions

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions.

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.

Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)

Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0

Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3

Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.2. The radicands of even-order roots can't be negative.Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.b. f(x) = 2x + 6

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.2. The radicands of even-order roots can't be negative.Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.b. f(x) = 2x + 6We can only extract square roots of nonnegative numbers

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.2. The radicands of even-order roots can't be negative.Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.b. f(x) = 2x + 6We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3.

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.2. The radicands of even-order roots can't be negative.Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.b. f(x) = 2x + 6We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3.So the domain = {all numbers x ≥ –3}.

Domain of FunctionsThere are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.2. The radicands of even-order roots can't be negative.Example C. Find the domain of the following functions.a. f(x) = 1/(2x + 6)The denominator can’t be 0 i.e. 2x + 6 = 0 x = –3So the domain = {all numbers except x = –3}.b. f(x) = 2x + 6We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0 x ≥ –3.So the domain = {all numbers x ≥ –3}. The requirement of having “the radicands ≥ 0” applies to the 4th root, or the 6th root, or any even-order root.

Domain of Functionsc. Find the domain of f(x) = 1 x + 6

– 1 4

Domain of Functionsc. Find the domain of f(x) =

The radicand of a 4th-root must be nonnegative.

1 x + 6 – 1 4

1 x + 6 – 1

Hence x must satisfy the inequality 1x + 6 – 1 > 0

1 x + 6 – 1

This is a sign-charting problem.

1 x + 6 – 1

This is a sign-charting problem. Put it in factored form: -x – 5

x + 6 > 0

1 x + 6 – 1

This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0

Draw the real line and sample points,

– 5– 6

1 x + 6 – 1

x + 6 > 0

Draw the real line and sample points, we get :

– 5– 6

UDF – – + – –

1 x + 6 – 1

x + 6 > 0

Draw the real line and sample points, we get :

– 5– 6

UDF – – + – –

The domain consists of the non–negative portioni.e. the domain is the interval (–6 ,–5].

Algebra of FunctionsWe may form expressions using functions or use expressions as inputs for functions.

c. f(2a) – g(a + b)

Algebra of FunctionsWe may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4.a. f(3) – g(4)

b. f(a+b)

c. f(2a) – g(a + b)

Algebra of FunctionsWe may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4.a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6,

b. f(a+b)

c. f(2a) – g(a + b)

Algebra of FunctionsWe may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4.a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8,

b. f(a+b)

c. f(2a) – g(a + b)

Algebra of FunctionsWe may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4.a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2.

b. f(a+b)

c. f(2a) – g(a + b)

b. f(a+b) = (a+b)2 – 2(a+b) + 3

c. f(2a) – g(a + b)

b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3

c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4]

Algebra of Functions

insert [ ] for subtraction

We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4.a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2.

b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3

c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4

b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3

c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4 = 4a2 – 7a – 3b + 7

b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3

Notation and Algebra of FunctionsOne important function-expression for any f(x) is its

"difference quotient": h

f(x+h) – f(x)

where h is another variable.

hf(x+h) – f(x)

Example E. Let f(x) = 2x2 – 3x + 4, simplify

hf(x+h) – f(x)

One important function-expression for any f(x) is its

"difference quotient":

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4]

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4]

= h2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4]

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4]

= h2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4]

= h4xh + 2h2 – 3h

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4]

= h2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4]

= h4xh + 2h2 – 3h

= hh(4x + 2h – 3)

hf(x+h) – f(x)

= h2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4]

= h2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4]

= h4xh + 2h2 – 3h

= hh(4x + 2h – 3)

= 4x + 2h – 3

Composition of FunctionsWhen one function is used as the input of another function, the outcome is called their composition.

Composition of FunctionsWhen one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).

Composition of FunctionsWhen one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f.

Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify

a. (f ○ g)(2)

a. (f ○ g)(2) = f(g(2))

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5)

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20

Composition of FunctionsWhen one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2)

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2))

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1c. (g ○ f)(x)

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3

a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3 = –12x + 23

In general, (f○g)(x) = (g○f)(x).

The Basic Language of FunctionsExercise G. Given the functionsf, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so.

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1) b. (f○g○h)(1)

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1)

(h○g○f)(–1) = h(g(f(–1))

b. (f○g○h)(1)

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))

b. (f○g○h)(1)

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3)

b. (f○g○h)(1)

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3) ≈ 4

b. (f○g○h)(1)

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1) b. (f○g○h)(1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3) ≈ 4

(f○g○h)(1) = f(g(h(1))

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1) b. (f○g○h)(1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3) ≈ 4

(f○g○h)(1) = f(g(h(1))

(1, ½ )

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1) b. (f○g○h)(1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3) ≈ 4

(f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)),

(1, ½ )

x y = g(x) –1 42 35 –36 47 2

y = h(x)

f(x) = 3x + 8

a. (h○g○f)(–1) b. (f○g○h)(1)

(h○g○f)(–1) = h(g(f(–1))= h(g(5))= h(–3) ≈ 4

(f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)),this is UDF because x = ½ in not in the domain of g.

(1, ½ )

Composition of FunctionsEquations may be posed in the language of functions.

Composition of FunctionsEquations may be posed in the language of functions.Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1,solve for x wherea. f(x) = 2

b. f(x) = g(x)