For the function f(x) = 2x2 – 3x + 4, the f(x) is the output and the 2x2 – 3x + 4 is the definition. To evaluate the function f(x) for an input of 5, we substitute 5 for x in the definition and simplify following the order of operations:

f(x) = 2x² - 3x + 4
f(5) = 2(5)² - 3(5) + 4
f(5) = 2(25) - 15 + 4
f(5) = 50 - 15 + 4
f(5) = 35 + 4
f(5) = 39

Using your calculator, enter , 2(5)² - 3(5) + 4
then press [ENTER]. Be careful that you use the parentheses correctly. Only the input number goes inside the parentheses. Coefficients and exponents go outside.

Arithmetic Operations on Functions: If f(x) and g(x) both exist, we define the following operations:

Addition: (f + g)(x) = f(x) + g(x) : add the definitions together by combining like terms and simplifying.

Subtraction:(f - g)(x) = f(x) - g(x) : distribute the negative to the definition of g(x)) {change the signs of ALL the terms in g(x)}, then combine like terms and simplify.

Multiplication: (f / g)(x) = f(x) • g(x) : distribute the terms of f(x) to the terms of g(x), then combine like terms and simplify.

Division: (f + g)(x) = f(x) / g(x) : the definition of f(x) is the numerator and the definition of g(x) is the denominator. Reduce any common factor of the entire numerator and denominator to 1. Do NOT reduce a term with a factor!

The ORDER of the functions is important when performing operations on functions. The functions MUST be worked in the order they are given, especially for subtraction and division.

Examples

For f(x) = 2x + 3 and g(x) = 4x – 5, find the following:

(f + g)(x) = f(x) + g(x) = 2x + 3 + 4x - 5 = 6x - 2
(f + g)(7) = f(7) + g(7) = (2)(7) + 3 + 4(7) - 5  Use this line with parentheses in calculator
= 14 + 3 + 28 - 5
= 45 - 5
= 40

 

(f - g)(x) = f(x) - g(x) = 2x + 3 - (4x - 5) = 2x + 3 - 4x + 5 = 2x + 8
(f - g)(7) = f(7) - g(7) = 2(7) + 3 - (4(7) - 5)
= 14 + 3 - (28 -5)
= 17 -23
= -6

 

(fg)(x) = f(x) • g(x) = (2x + 3)(4x – 5) = 8x2 – 10x +12x – 15 = 8x2 + 2x – 15
(fg)(7) = f(7) • g(7) = (2(7) + 3) • (4(7) - 5)
= (14 + 3) • (28 - 5)
= 17 • 23
= 391

 

(f / g)(x) = f(x) / g(x) = 2x + 3 / 4x - 5
(Since there are no common factors of the numerator and denominator ( the 2x and 4x are terms), the fraction can not be reduced.)
(f / g)(7) = f(7) / g(x) = 2(7) + 3 / 4(7) - 5
= (14 + 3) / (28 - 5)
= 17 / 23

For your calculator do this: (2(7)+3)/(4(7)–5) [MATH][ 1 ][ENTER]

Another concept is called Composition of Functions. This is a model of performing a sequence of tasks.

To find the Composite function f of g(x): (f • g)(x) = f(g(x)) : reads as “the f of the g of x.” We substitute the definition of g(x) for the x in the definition of f(x). When evaluating for a real number input, we find the output of g(x), then use this output for the input of f(x). Evaluate the definition of g(x) for the given number, then take the number you get and use it in the definition of f(x).

Examples

For f(x) = 2x + 3 and g(x) = 4x – 5, find the following:

(f • g)(7)

= f(g(7))  = f(4(7) - 5)  = f(28 - 5)  = f(23)  = f(23) + 3  = 46 + 3  = 49

and

(g • f)(7)

= g(f(7)  = g(2(7) + 3)  = g(14 + 3)  = g(17)  = 4(17) - 5  = 68 - 5  - 63

The correct order in the sequence is vital when we compose functions. The real number you are given is the starting point used in the function closest to it. The output of that function becomes the input in the other function.

Inverse Operations: addition and subtraction are inverse operations as are multiplication and division: one operation undoes the other.

ƒ –1(x): the inverse of the function f(x). Read as “the f inverse of x.” In this case the –1 is NOT an exponent but means the inverse function. ONLY 1-to-1 functions have inverse functions.

One-to-one function: In a 1-to-1 function, each input (x) must be paired to 1 and only 1 output (f(x)). This means that no two x’s will generate the same f(x). If different inputs produce the same output, then an inverse function does not exist. A 1-to-1 function passes first the Vertical line test (to be a function) and then the Horizontal line test (to be 1-to-1).

Recognizing 1-to-1 functions: Some functions are always 1-to-1 , some are never 1-to-1, and some you have to check.

1. A linear function of the type f(x) = mx + b is ALWAYS 1-to-1. The definition side (right of the =) MUST contain an x-term.
2. A linear function of the type f(x) = b, where b is a real number constant, are NEVER 1-to-1 since the graph is a horizontal line and can not pass the horizontal line test.
3. Passes both Vertical and Horizontal line tests: Vertical line test: no vertical line can be drawn anywhere through the graph of the function and intersect more than once. This is a test to see if the relation is a function. Then, to be 1-to-1, the function must also pass the Horizontal line test: no horizontal line can be drawn anywhere through the graph of the function and intersect more than once.
4. Quadratic functions are NEVER 1-to-1, nor any function having an even (2, 4, 6, 8, . . .) exponent as the largest exponent. WHEN the domain is restricted, the partial graph may be 1-to-1.
5. Some Cubic (biggest exponent = 3) functions are 1-to-1, some are not. Cubic functions of the type f(x) = x3 and f(x) = x3 + d, where d is a real constant, are 1-to-1. If the cubic has another x-term, check it by graphing. (see example 3 on page 388)

To find f–1(x): let f(x) = x3 – 2

First: Is f(x) a 1-to-1 function? Yes, so proceed.

Write the function in equations form: y = x3 – 2

Invert (swap) ONLY the variables: x = y3 – 2

Solve for y : Add 2 to both sides: x + 2 = y3 – 2 + 2

Simplify: x + 2 = y3

Take the third root of both sides: 3√x+2 = 3√y³

Simplify: 3√x + 2 = y

Rewrite in function notation: ƒ¹(x) = 3√x + 2

Verify (2 parts):

Part 1: If the inverse is correct, then (ƒ¹  • ƒ(x)) = x

Replace x inƒ–1(x) withƒ(x): (ƒ¹  • ƒ(x)) = 3√x³ + 2 - 2

Combine like terms under the radical: (ƒ¹  • ƒ(x)) = 3√x³

Take the cube root:(ƒ¹  • ƒ(x)) = x

Part 1 is verified.

Part 2: If the inverse is correct, then (ƒ • ƒ^-1(x)) = x

Replace x inƒ(x) withƒ–1(x): (ƒ • ƒ^-1(x)) = (3√x + 2)³ - 2

Raise to the third power: (ƒ • ƒ^-1(x)) = (x + 2 - 2

Combine like terms: (ƒ • ƒ^-1(x)) = x

Part 2 is verified.

An exponential function ƒ with base a and coefficient C is represented by:
ƒ(x) = Ca^x, a > 0, a ≠ 1, and C > 0.

Note: Neither the coefficient C nor the base a can be negative or zero! They MUST be positive real numbers.

To determine whether a function is linear or exponential from tabular data:

Linear data: (Table 5.14, p. 402)

x	0	1	2	3	4	5
y = g(x)	3	5	7	9	11	13

When data are linear, the difference between the x’s is always the same number and the difference between the y’s is always the same number (added or subtracted). Here, the difference in x’s is always 1 and the difference in y’s is always 2, so the data are linear. To find the function, we need to know the slope (m) and the y-coordinate of the y-intercept {the b from the point (0, b)}.
Remember: the slope is the difference in y’s (Rise) divided by the difference in x’s (Run). m = 2/1 = 2. To find b, locate 0 in the x-row and use the y-value under it: when x = 0, y = 3, so b = 3.
Now write in linear function form: g(x) = mx + b, so g(x) = 2x + 3

Exponential data: (Table 5.15, p. 402)

x	0	1	2	3	4	5
y = ƒ(x)	3	6	12	24	48	96

 

When data are exponential, the difference between the x’s will probably be the same number BUT the difference between the y’s changes exponentially (multiplication). Here, the difference in x’s is always 1 but the difference in y’s is always multiplied by 2 to a power, so the data are exponential. To find the function, we need to know the base (a) and the coefficient (C), which is the y-coordinate of the y-intercept {the b from the point (0, b)}.

To find b, locate 0 in the x-row and use the y-value under it: when x = 0, y = 3, so C = 3. To find a, find out what C is multiplied by from one value of y to another: from 3 to 6 is 3 * 2, from six to 12 is 6 * 2, from 12 to 24 is 12*2, from 24 to 48 is 24*2, from 48 to 96 is 48*2. Since we get the next y-value by multiplying the previous y-value times 2, 2 is our base: a = 2.

Now write in exponential function form: f(x) = Cax, so f(x) = 3(2)x

Natural Exponential Function: ƒ(x) = ex, where e ~ 2.718282 (~ means approximately). e is the "natural number" and you'll see it again in the continuously compounded interest formula and in natural logarithms.

Two formulas for Compound Interest:

when compounded periodically

A = P(1 +(r/n))nt

A is the Amount after compounding, P is the Principal, r is the interest Rate in decimal form, n is the Number of times the interest is paid in one year, and t is the Time in years.

Example: What will be the total amount repaid from a $2000 loan @ 10% compounded monthly for 4 years?

Use your calculator and enter: 2000(1 + ( .1/12))^(12*4) [ENTER]

answer: $2978.71

when compounded continuously

A = Pert

A is the Amount after compounding, P is the Principal, r is the interest Rate in decimal form, and t is the Time in years.

Example: What will be the total amount repaid from a $2000 loan @ 10% compounded continuously for 4 years?

Use your calculator and enter: 2000e^(.1*4) [ENTER] e on the calculator is [2nd] [ ÷ ]

answer: $2983.65

Logarithmic Functions and Models

A logarithm is an exponent.

Two types of logarithms:
1) common log (base 10), written as log, read as log or common log of
Example: log 1000 is read as the common log of 1000
Asks the question: what exponent raises 10 to a product of 1000?
Since 103 = 1000, then log 1000 = 3.
In this example, 10 is the base, 3 is the exponent and so is log 1000, and 1000 is the product.

2) natural log (base e) written ln, read as natural log of
Example: ln 1097 is read as the natural log of 1097
Asks the question: what exponent raises e to a product of 1097?
Since e7.00033446 = 1097, then ln 1097 = 7.00033446.
In this example, e is the base, both 7.00033446 and ln 1097 represent the exponent, and 1097 is the product.

The only thing new about logarithms is how the information is written. You need to practice so that you can identify the parts (base, product, and exponent) and where they appear in the pattern.

Sometimes we need to use a base other than 10 or e, for example 5. Then we write log5 125 = 3. In this case, the base is 5, the exponent is represented by both 3 and log5 125, and the product is 125. To check, we see if 53 = 125 is true. Since it is, then we know that log5 125 = 3 is correct.

Change of base pattern: The graphing calculator has only a log key and a ln key. If we need to find a logarithm for a different base, we need to know how the use the change of base pattern. The pattern is the log of the product divided by the log of the new(different) base: log(product) / log (newbase)

To find log5 78125, we would need to change from base 10 (common log) to base 5. We need to find: log(78125) / log(5)

On the calculator, press [log] shows log(
Type 78125 then a close parenthesis shows log(78125)
Now press the division key shows log(78125)/
Press the log key again shows log(78125)/log(
Type the new base, close parenthesis shows log(78125)/log(5)
Press enter, shows 7. This is correct because 57 = 78125.

What is called newbase above is really any base that is diferent from the common base 10 (log) and the natural base e (ln)