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Question 739462: What is the range of :h(x)=2x+1/x-3
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! What you wrote, h(x)=2x+1/x-3, is  , 
But maybe you meant h(x)=(2x+1)/(x-3), which is  , 
Both functions get as low as 
and as high as  ,
but we have to figure out if there are some in-between values that the functions cannot achieve.
I will help you discover what each of those functions does.
EASIEST FIRST:
When  has a very large absolute value, the last term gets very small and the value of the function gets very close to  .
You can get as close as you want, by making that last term as small (in absolute value) as necessary.
So the line  is a horizontal asymptote.
However,  can not equal , because the last term can never be zero.
So you could say that the range of that function is all the real numbers, except 2.
THE OTHER FUNCTION:
does something different.
When has a very large absolute value, the term  gets very small and the value of the function gets very close to the value of  .
You can get as close as you want, by making that term as small in absolute value as necessary.
So the line is an oblique asymptote.
For  ,  and
for  , 
so the function never takes the value  , but there are other forbidden values.
You can figure out that cannot take any values between  and  and that means that
THE RANGE IS all the real numbers  such that
 or 
You may want to express that in whatever notation is expected of you,
maybe as a union of intervals:
(-infinity,-3-2sqrt(2)] U [-3+2sqrt(2),infinity)
If you know about derivatives, you would find that the derivative of  is
h'(x)= , which has zeros at  and 
That means that the function will have a maximum with at
and a minimum with at
You could calculate them as
 and
Without mentioning derivatives, you could figure out what values of
 are possible for some real number by solving that equation for
-->  -->  --> 
You know that there will be one or two solutions wherever the discriminant is not negative, meaning when
 -->  --> 
The solutions to are  and
the solution for is the range of :
and
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