Functions

(1)  Find f(2) when f(x) = x³ - 5
                             f(2) = 2³ - 5 = 8 - 5 = 3

(2)  For f(x) = x² + 3,  find    f(x + h) -  f(x)
                                                         h
        

Domain:  x-values being used to construct the function

Range:  y-values being used to construct the function

Examples:    Consider the domain and range to be values of the set of Real Numbers.

1.  g(x) = __3__
              x² - 1

Graph g(x) in your calculator.

Domain:  The denominator cannot be zero.  So x² - 1 = 0 cannot be.
                                                                       (x - 1)(x + 1) = 0
                                                                          x = 1, -1
     So the domain is all reals except x = 1, -1.

Range:  All real numbers except y = 0  because __3__ can never be 0.
                                                                       x² - 1
Zeros:  None, because there are no solutions to the equation __1__ = 0.
                                                                                           x² - 1

         g(2):  g(2)  =  __3__   =   3/3   =   1
                           2² - 1

2.  f(x) = |x² - 9|

Again, graph f(x) in your calculator.

Domain:  Since  x² - 9  can be any value, then x itself can take on any value. Answer:  All real numbers

Range:  y > 0 because f(x) is an absolute value which is always positive.

Zeros:  x = -3,3

g(2):  |2² - 9|   =  |4 - 9|   =   |-5|   =   5
 

Again, graph h(x) in your calculator.



                  Window:  -10 < x < 10  and  -10 < y < 10
                    Zoom 6

Domain:  All Real Numbers, because all possible x-values are listed in the intervals on the right

Range:  Inspecting the graph, we wee that   y < -5 and  -4 < y < 5

Zeros:  Inspect the graph for the places it actually touches the x-axis:   x = -2, 2

Problems

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