Unit 2 Function Algebra

Range

The set of all possible output values for a function is called the range.

Gravitational Acceleration on Earth

9.8 m/s^2

Composition

A combination of two functions such that the output from the first function be omes teh input for the second is called a composition.

One-to-one Function Check Without Graph

A function, f, is one-to-one if f(x1)=f(x2) implies x1=x2. Check for f(x)=5x-1. First, substitute two different variables into the function. f(x1)=5x1-1 and f(x2)=5x2-1 Next, set the rules for the functions equal to each other and isolate the variables. 5x1-1=5x2-1 5x1-1+1=5x2-1+1 5x1/5=5x2/5 x1=x2 It is one-to-one. Check for h(x)=x^2 h(x1)=x1^2 and h(x2)=x2^2 x1^2=x2^2 sqrtx1^2=sqrtx2^2 +-x1=+-x2 Since x1 could be positive when x2 is negative or vice versa, x1 and x2 are not necessarily equal. Therefore, h(x) is not a one-to-one function.

Set Builder Notation

A method of writing a set of numbers by describing the properties of its members. Set builder notation is written in braces. After the left brace, you name the variable you are describing, then put a vertical bar, and then write a mathematical description of the set of numbers represented by the given variable. A domain that is between but not including 2 and 5 would be written as {x | 2 < x <5} in set builder notation. This is read "the set of all x such that 2 is less than x and x is less than 5."

Open and Closed Dots

An open dot rperesents a point that is not included in the function, while a closed dot represents point that is included in the function.

Types of Intervals

An open interval is where the endpoints are not included. It is indicated using parentheses. A closed interval is an interval where are included. It is indicated using brackets. An interval where one of the endpoints is included and one is not included is called a half-open interval. It is indicated by using a parenthesis on the open side and a bracket on the closed side. Use the positive or negative infinity symbol, for a domain or range that continues in one or both directions continuously. Always use parentheses. An interval that starts, ends, or starts and ends at infinity is called a non-ending interval.

Restricting the Domain Part 2

Consider the function f(x)=(x-3)^2+2. This function is not one-to-one. You will need to first restrict the domain to find the inverse. This function is a translation of the function g(x)=x^2. The graph is shifted to the right 3 units and up 2 units. The vertex of the parabola is (3,2). This is the turning point of the parabola; therefore, it is at the x-value of this point that the restricted domain must end or begin. You can choose either side of the parabola. If you restrict the domain to (-infinity,3], you create a function consisting of the left half ot he parabola. This function is now one-to-one and will have an inverse function. Find the inverse function algebraically by interchanging x and y and then solving for y. y=(x-3)^2+2 Replace f(x) with y. x=(y-3)^2+2 Interchange x and y. x-2=(y-3)^2 Subtract 2 from both sides of the equation. +-sqrtx-2=y-3 Take the square root of both sides of the equation. y=3+-sqrtx-2 Add 3 to both sides of the equation. f^-1(x)=3+-sqrtx-2 Replace y with f^-1(x) You can't have one y-value paired with two x-values, so you use y=3-sqrtx-2 since the range of f^-1(x) must be (-infinity,3]. So, f^-1(x)=3-sqrtx-2. You can graph the function and its inverse to check your work or to help identify the domain and range.

Graphs of Inverses

Consider the function f(x)=6x+1. This is a linear function. It is also one-to-one. Therefore, its inverse is a function. Find the inverse function. y=6x+1 Replace f(x) with y. x=6y+1 Interchange x and y. x-1=6y Subtract 1 from both sides of the equation. y=x-1/6 Divide both sides of the equation by 6. Therefore, f^-1(x)=x-1/6. Now, graph both of these functions. Interchanging the coordinates of a point on the original function's graph produces an ordered pair of coordinates of a point that is on the inverse function's graph. For example, some points on the graph of the original function, f(x), such as (-1,-5), (0,1), and (1,7). The points with these coordinates interchanged--(-5,-1), (1,0), and (7,1)--lie on the graph of the inverse function.

Graphs of Inverses Part 2

Consider the function g(x)=x^3. This is a one-to-one function, and it therefore has an inverse function. y=x^3 Replace f(x) with y. x=y^3 Interchange x and y. y=^3sqrtx Take the cube root of both sides of the equation. The inverse function is g^-1(x)=^3sqrtx. Graph both of these functions. Again, the graph shows that by interchanging the coordinates of the points from the original function, you get the points that lie on the graph of the inverse function.

Restricting the Domain

Consider the quadratic parent function, f(x)=x^2. From its graph, you can see that this function is not one-to-one, as it does not pass the horizontal line test. However, by restricting the domain of the function, you can create a function that is one-to-one. For example, the graph of g(x)-x^2 for the domain [0,infinity). This function is one-to-one. Therefore, it will have an inverse function. The inverse function will have a range of [0,infinity). You can find the inverse function either algebraically or graphically. y=x^2 Replace g(x) with y. x=y^2 Interchange x and y. y=+-sqrtx,x> or equal to 0 Take the square root of both sides of the equation. You can't have one y-value paired with two x-values, so you can't have both +sqrtx and -sqrtx. Since the range of the inverse is [0,infinity), g^-1(x)=sqrtx, x> or equal to 0.

Square Roots

Example: f(d)sqrt9.8d You can take the square root of any real number, a, such that a greater than or equal to 0. The square root of a negative number does not result in a real number, so that will not be part of the domain. Create an inequality such that the radicand, the expression under the radical, is greater than or equal to 0, and solve to determine which input values will result in a real number output. 9.8d greater than or equal to 0 Divide both sides of the inequality by 9.8. d greater than or equal to 0 The domain is any real number greater than or equal to 0. So the domain of f is [0, infinite) or {x|x greater than or equal to 0}.

Second Method

Find the domain of (g o f)(x) when g(x)=1/x+1 and f(x)=2/x-4. First, find the domain of the inner function, which in this case is f(x). The domain of f(x) is (-infinite, 4) U(4, inifinite) since x cannot be 4. Second method You know the domain of the outer function, g(x)=1/x+1, cannot contain -1. Therefore, the inner function cannot have an output of -1. Set the inner function, f(x), equal to -1 to find any values that need to be restricted from the domain. 2/x-4=-1 2=-1(x-4) Multiply both sides of the equation by (x-4). 2=-x+4 Apply the Distributive Property. -2=-x Subtract 4 from both sides of the equation. x=2 Divide both sides of the equation by -1. Thus, 2 must be restricted from the domain. Therefore, the domain for the composite function cannot contain 2 or 4. So, the second method also shows that the domain of the composite function is (-infinite, 2)U(2,4)U(4, infinite).

Horizontal Line Test

For any function, you can use the horizontal line test ot make sure that every y-value is paired with one and only one x-value.

Find Domain (and Range) with Restrictions

For f(x)=sqrtx-5 and g(x)=1/x+4, find the function (f+g)(x) and determine its domain and range. (f+g)(x)=f(x)+g(x) (f+g)(x)=sqrtx-5+1/x+4 There is no need to simplify this function further. To find the domain of the new function, first find the domain of each of the original functions. The domain of f(x) is [5, infinite) because x-5 must be greater than or equal to 0 to produce a real-number output. The domain of g(x) is (-infinite, -4) U (-4, infinite) because x+4 cannot equal 0. The domain of (f+g)(x) must include both restrictions, so the domain is all real numbers greater than or equal to 5 and not equal to -4. Since the restriction greater than 5 already rules out -4 as an input-value, the domain of (f+g)(x) is [5, infinite). Since the domain is [5, infinite), evaluate (f+g)(5) to determine the minimum value of the function. (f+g)(5)=sqrt5-5 +1/5+4 (f+g)(5)=1/9 Therefore, the range is [1/9, infinite).

One-to-one Functions

Functions in which each element in the range is associated with exactly one element in the domain. For a relation to be a one-to-one function, each element in the domain must map to one, and only one, element in the range, and each element in the range must map to one, and only one, element in the domain.

Adding and Subtracting Functions

Given any two functions f(x) and g(x), you can add or subtract the functions by adding or subtracting their rules. The notation for the sum of functions f(x) and g(x) is (f+g)(x).(f+g)(x)=f(x)+g(x) The notation for the difference between f(x) and g(x) is (f-g)(x). (f-g)(x)=f(x)-g(x)

Division Example

Given f(x)=2x^2-3x+4 and g(x)=x+3 (f divided by g)(x) When finding the quotient of two functions f and g, write the functions in this format: f(x)/g(x). Then substitute using the given functions: 2x^2-3x+4/x+3. Since the numerator is not factorable, leave the quotient in its current format. The solution is 2x^2-3x+4/x+3.

Multiplication Example

Given f(x)=2x^2-3x+4 and g(x)=x+3 (f*g)(x) When finding the product of functions f and g, write the two functions in this format: f(x)*g(x). Then substitute using the given functions: 2x^2-3x+4(x+3). Now carry out the multiplication by simplifying via distribution: 2x^3-3x^2+4x+6x^2-9x+12. Next, combine like terms for a final solution: 2x^3+3x^2-5x+12.

Composite Functions Percentages

If a sale offers a 25% discount, the function g(x)=0.75x represents the 25% discount because you are only playing 75% of the original price. The function f(x)=0.95 represents the store credit card discount because you are only paying 95% of the original price. Since the discounts are applied one at a time, you can write a composition of functions so that you calculate the first discount and then fin dhte second discount on the result of that first discount. Assume that the 25% discount is calculated first. You would first find g(x) and then evaluate f(x) for the result. Composition of functions allows you to do this in one step. Since you want to find g(x) first, it becomes the inner function in the composition. So in this case, find (f o g)(x). (f o g)(x)=f(g(x)) (f o g)(x)=f(0.75x) Replace g(x) with its rule. (f o g)(x)=0.95(0.75x) Substitute 0.75x for x in f(x). (f o g)(x)=0.7125x Simplify the new function. You can now see that you are paying 71.25% of the original price after both discounts. Since the original price was $50.00, you can evaluate the composite function for x=50 to find the sale price. (f o g)(x)=0.7125(50 (f o g)(x)=35.625 The sale price is 35.63. You could cehck this by evaluating g(x) for x=50, and then evaluating f(x) for the result.

Compose Functions Algebraically (f o g)

Let f(x)=x+3 and g(x)=2x. Find f(g(2)). By definition, this means to evaluate g(2) and then substitute the output into f(x). First find g(2). g(x)=2x g(2)=2(2) g(2)=4 Now substitute 4 into f(x). f(x)=x+3 f(4)=4+3 f(4)=7 Therefore, f(g(2))=7. Instead of doing these seaprately, you can find the composite function algebraically by substituting the rule for g(x) for x in f(x). (f o g)(x)=f(g(x)) (f o g)(x)=f(2x) Replace g(x) with its rule. (f o g)(x)=2x+3 Substitute the rule for g(x) for x in f(x). Once you have the composite function you can substitute the input value into the omposite function. (f o g)(2)=2(2)+3 (f o g)(2)=7

Interval Notation

Method of writing a set of numbers where parentheses are used for an exclusive element and a bracket is used for an inclusive element. It is written as a pair of numbers when the number on the left is the smallest bound, while the number on the right is the largest bound.

Multiplying and Dividing Function

Multiplying and dividing functions is similar to adding and subtracting; you perform the operation on the function rules to make a new function. To find the product of two functions, multiply their rules. (f*g)(x)=f(x)*g(x) To find the quotient of two functions, divide their rules. Restrict the domain as necessary so that the divisor is not 0. (f/g)(x)=f(x)/g(x), g(x)is not equal to 0. To find the domain and range of the resulting function, begin by finding the domain of the original functions. The domain of the resulting function is ther intersection of the domains of the original functions. There is one extra consideratin when you are dealing with division of function. g(x) cannot be equal to 0, so you must restrict the domain of the resulting ufnction accordingly. Once you have the domain, graph the function over that domain to find the range.

Domain and Range of a Function and Its Inverse

Since the range of a function is the domain of its inverse, you can find the range of a function by finding the domain of its inverse function. Similarly, you can find the range of the inverse function by finding the domain of the original function. Find the domain and range for the one-to-one function f(x)=x+1/x-2 and its inverse. First, find the domain of f(x). The domain of f(x) cannot include 2, and is therefore (-infinity,2)U(2,infinity). Since the range of the inverse function is the same as the domain of the original function, the range of f^-1 is also (-infinity,2)U(2,infinity). Next, find the inverse function. Replace f(x) with y. y=x+1/x-2 Interchange x and y. x=y+1/y-2 Multiply both sides of the equation by y-2. x(y-2)=y+1 Apply the Distributive Property xy-2x=y+1 Add 2x and -y to both sides of the equation. xy-y=2x+1 Factor out the y from xy-y. y(x-1)=2x+1 Divide both sides of the equation by x-1. y=2x+1/x-1 Replace y with f^-1(x). f^-1(x)=2x+1/x-1 The domain of f^-1(x) cannot include 1, so it is (-infinity,1)U(1,infinity). Since the range of f(x) is the same as the domain of its inverse, the range of f(x) is also (-infinity,1)U(1,infinity).

Inverse Function

The inverse function will produce the same ordered pairs as the original function, but their coordinates will be interchanged. A function, f, has an inverse function if it is one-to-one. The inverse of f(x) is written as f^-1(x), which is read as "the inverse of f of x." The -1 in this notation is not an exponent; it is simply part of the notation used to indicate the inverse of a function.

Composing Inverse Functions

The inverse of a function, when composed with the original function, undoes the original function, just like an inverse operation undoes the original operation performed. Consider a function, f(x), that maps 6 onto 5. By definition, the inverse function f^-1(x) mpas 5 onto 6. If you substitute 6 into the original function and then substitute the output 5 into the inverse function, you would get back the original value, 6, and vice versa. Conversely, evaluating a second function for the output of a first function is what is accomplished by using a composite function. In other words, for functin f(x) and its inverse f^-1(x), (f o f^-1)(x)=x and (f^-1 o f)(x)=x. The output of the composed function is the same value as the input, no matter which function is evaluated first. If (f o g)(x)=x and (g o f)(x)=x, then f(x) and g(x) are inverses of each other.

Domain

The set of all possible input values for a function is called the domain.

Domain of Composite Functions

To determine the domain of a composite function, you must first determine the domain for the inner function, as that is the function that is performed first. If a value is not in the domain of the inner function, it cannot be in the domain of the composite function. Next, you must consider the domain of the outer function using one of two methods. In thee first method, you simplify the rule for the composite function and check to see if there are any furhter restricitons on the domain. In the second method, you determine what the outer function requires of the range of the inner function. In other words, you determine if there are output values that the inner function might produce that cannot be input into the outer function, and exclude the domain from the input values that map to those output values. Find the domain of the composite function (f o g)(x) when f(x)=1/x and g(x)=sqrtx. First, find the domain of the inner function, which in this case is g(x). The domain of g(x) is [0, infinite). Using the first method, determine the domain restricitions of (f o g)(x) by simplifying the rule for the composition of the functions. (f o g)(x)=f(g(x)) (f o g)(x)=f(sqrtx) (f o g)(x)=1/sqrtx Substitute the rule for g(x) for x in f(x). The denominator cannot be 0, so sqrtx cannot be 0. Therefore, the domain of (f o g)(x) is (0, infinite). Using the second method, you see that the restriction on the domain of f(x) is that x cannot be 0. So, without finding the composite function, you know that the output of the innner function g(x) cannot be eqaul to because 0 cannot be input into f(x). Find the input that makes g(x) equal to 0 and restrict it from the domain. sqrtx=0 x=0 Restrict 0 from the domain. Again, you see that the domain of (f o g)(x) is (0, infinite).

Composite Functions

To find the composition of the functions f(x) and g(x), you can find each output value of g(x) and then use them as input values for f(x). The notation f(g(x)) shows that you are substituting the values of the inner function g(x) into f(x). Mathematicians use the notation (f o g)(x), which means the same thing as f(g(x)), to represent composite functions. The function (f o g)(x) is read as "f composed with g of x." In this notation, the second variable represents the inner function. So (f o g)(x)=f(g(x)) and (g o f)(x)=g(f(x)).

Combination or Union of Intervals

When a domain or range is not continuous, write an interval for each continuous section of the domain or range. To indicate that the domain or range is a combination, or union, of the intervals, use the U shaped symbol.

Domain from an Equation

You can cube and square any real number. You can also multiply by or add to any real number. This tells you that the x can be any real number, or that the domain is (-infinite, infinite). This logic will hold for any polynomial. Tehrefore, the domain of any polynomial is the set of all real numbers. To find waht input value cannot be in the domain, set the expression in the denominator equal to 0 and solve. g(x)=1/x-7 x-7=0 x=7 For g(7), the function is undefined, so the domain is all real numbers except for 7.

Are the functions inverses?

f(x)=2/7x-4 g(x)=7/2x+14 (f o g)(x) (f o g)=f(g(x)) f(g(x))=2/7(7/2x+14)-4 =2/7(7/2x)+2/7(14)-4 =x+28/7-4 =x+4-4 =x Yes (g o f)=g(f(x)) g(f(x))=7/2(2/7x-4)+14 =7/2(2/7x)-7/2(4)+14 =x-28/2+14 =x-14+14 =x Yes

Composite of Functions

g(x)=x+3 and f(x)=2x-4 What is f(g(x))? f(g(x))=2(x+3)-4 =2x+6-4 =2x+2

Velocity of an Object Orbiting Another Object

v(r)=sqrt(G*Mcentral/r), where G is the universal gravitational constant, Mcentral is the mass of the object around which the object is orbiting, and r is the radius of the orbit. The universal gravitational constant is 6.673*10^-11 (Nm^2/kg^2). For a satellite orbiting Earth, the mass of Earth 5.9726*10^24 kg, is Mcentral. Therefore, you can find the veloctiy, v(r), of a satellite orbiting Earth with a radius, r, using the function v(r)=sqrt(6.673*10^-11)(5.9726*10^24)/r. Note that r represents a distance, and since it is the denominator of a fraction, r>0. First, check to make sure that the function is one-to-one so that the inverse will also be a function. Set v(r1)=v(r2). sqrt(6.673*10^-11)(5.9726*10^24)/r1=sqrt(6.673*10^-11)(5.9726*10^24)/r2 Square both sides of the equation. (6.673*10^-11)(5.9726*10^24)/r1=(6.673*10^-11)(5.9726*10^24)/r2 Multiply both sides of the equation by r1r2. r2(6.673*10^-11)(5.9726*10^24)=r1(6.673*10^-11)(5.9726*10^24) Divide both sides of the equation by (6.673*10^-11)(5.9726*10^24). r1=r2. Next, use the steps for finding the inverse function. For simplificity, write the function in terms of the radius, x. v(x)=sqrt(6.673*10^-11)(5.9726*10^24)/x Replace v(x) with y. y=sqrt(6.673*10^-11)(5.9726*10^24)/x Interchange x and y. x=sqrt(6.673*10^-11)(5.9726*10^24)/y Square both sides of the equation. x^2=(6.673*10^-11)(5.9726*10^24)/y Multiply both sides of the equation by y.x^2y=(6.673*10^-11)(5.9726*10^24) Divide both sides of the equation by x^2. y=(6.673*10^-11)(5.9726*10^24)/x^2 Replace y with v^-1(x). v^-1(x)=(6.673*10^-11)(5.9726*10^24)/x^2 So, the inverse function is v^-1(x)=(6.673*10^-11)(5.9726*10^24)/x^2. Note that since x represents the velocity, and since it is squared in the denominator of a fraction, x>0.