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Finding the domain/range of a funtion help

Can someone explain step by step how to identify them?

For example:

f(x)= (x-4)^2 - 16 , where x is any real number and x>0
g(x)= 8/(1-x), where x is any real number and x<1

a) Find the range of f


Thank you :blushing:
Original post by Dinaa
Can someone explain step by step how to identify them?

For example:

f(x)= (x-4)^2 - 16 , where x is any real number and x>0
g(x)= 8/(1-x), where x is any real number and x<1

a) Find the range of f


Thank you :blushing:


For your first example, a graph may help (think about the minimum point).

For g(x), a graph may also help.

For a lot of these questions testing the values of f around the endpoints of the domain (eg something like 0.99 for g(x) as well as 'large' values (infinity, -infinity) may give you an idea of what the range is.
Reply 2
Original post by Dinaa
Can someone explain step by step how to identify them?

For example:

f(x)= (x-4)^2 - 16 , where x is any real number and x>0
g(x)= 8/(1-x), where x is any real number and x<1

a) Find the range of f


Thank you :blushing:


sketch just the shape of the graph for all reals
mark the coordinates of the minimum point
"throw" away the graph for which x<0
then you will see the graph lives in y for every value greater or equal to -16
Draw f(x) and look at the y values
Reply 4
Original post by SeanFM
For your first example, a graph may help (think about the minimum point).

For g(x), a graph may also help.

For a lot of these questions testing the values of f around the endpoints of the domain (eg something like 0.99 for g(x) as well as 'large' values (infinity, -infinity) may give you an idea of what the range is.


Original post by TeeEm
sketch just the shape of the graph for all reals
mark the coordinates of the minimum point
"throw" away the graph for which x<0
then you will see the graph lives in y for every value greater or equal to -16


I did draw it (may/probably) wrong because I got: when x=0, y=0.. so I assumed the range would be greater than 0.
Reply 5
Original post by Dinaa
I did draw it (may/probably) wrong because I got: when x=0, y=0.. so I assumed the range would be greater than 0.


that is true but there is a lowest value of y which is -16
Reply 6
Remember any number squared is positive. What value of x gives the smallest (x - 4)^2? There's your minimum y.

If x is strictly bigger than 0, x > 0, then x = 0 won't be on your graph.

Usually helps to think about extreme values. On g(x) think about x = 0.99999999999999999999999, which will give you a massive number. Then think x = -10^9, this will give you an extremely small number.
(edited 8 years ago)

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