Beginning_and_Intermediate_Algebra- Tyler Wallace

# World view note the concept of a function was first

• Notes
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World View Note: The concept of a function was first introduced by Arab mathematician Sharaf al-Din al-Tusi in the late 12th century Once we know a relationship is a function, we may be interested in what values can be put into the equations. The values that are put into an equation (generally the x values) are called the domain . When finding the domain, often it is easier to consider what cannot happen in a given function, then exclude those values. Example 507. Find the domain : f ( x )= 3 x 1 x 2 + x 6 With fractions , zero can t be in denominator x 2 + x 6 0 Solve by factoring ( x +3)( x 2) 0 Set each factor not equal to zero x +3 0 and x 2 0 Solve each equation 3 3 +2+2 x 3 , 2 Our Solution The notation in the previous example tells us that x can be any value except for 3 and 2 . If x were one of those two values, the function would be undefined. Example 508. Find the domain : f ( x )=3 x 2 x With this equation there are no bad values All Real Numbers or R Our Solution In the above example there are no real numbers that make the function unde- fined. This means any number can be used for x . Example 509. Find the domain : f ( x )= 2 x 3 Square roots can t be negative 388

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2 x 3 greaterorequalslant 0 Set up an inequality +3+3 Solve 2 x greaterorequalslant 3 2 2 x greaterorequalslant 3 2 Our Solution The notation in the above example states that our variable can be 3 2 or any number larger than 3 2 . But any number smaller would make the function unde- fined (without using imaginary numbers). Another use of function notation is to easily plug values into functions. If we want to substitute a variable for a value (or an expression) we simply replace the vari- able with what we want to plug in. This is shown in the following examples. Example 510. f ( x )=3 x 2 4 x ; find f ( 2) Substitute 2 in for x in the function f ( 2)=3( 2) 2 4( 2) Evaluate , exponents first f ( 2)=3(4) 4( 2) Multiply f ( 2)= 12 +8 Add f ( 2)= 20 Our Solution Example 511. h ( x )=3 2 x 6 ; find h (4) Substitute 4 in for x in the function h (4)=3 2(4) 6 Simplify exponent , mutiplying first h (4)=3 8 6 Subtract in exponent h (4)=3 2 Evaluate exponent h (4)=9 Our Solution Example 512. k ( a )=2 | a +4 | ; find k ( 7) Substitute 7 in for a in the function k ( 7)=2 | − 7+4 | Add inside absolute values k ( 7)=2 | − 3 | Evaluate absolute value k ( 7)=2(3) Multiply 389
k ( 7)=6 Our Solution As the above examples show, the function can take many different forms, but the pattern to evaluate the function is always the same, replace the variable with what is in parenthesis and simplify. We can also substitute expressions into func- tions using the same process. Often the expressions use the same variable, it is important to remember each variable is replaced by whatever is in parenthesis.

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