Question

Do we have a way of determine the image of a function with limitations, by not useing graph and how?
Fx. 2x^2-x-1/(x^2+3x+2), x∈[0,∞[

Answer 1

Well, you can just examine the function, no?

It is always increasing because of the 2x^2/x^2.

The only place anything funny is happening is between 0 and 1 which you can just quickly calculate.

Answer 2

Sorry ment (2x^2-x-1)/(x^2+3x+2), x∈[0,∞[

Answer 3

It's the same, no?

Answer 4

No becuase that function would have a lim for x-> infinity = 2

Answer 5

And ?

Answer 6

Your question is to describe the graph, is that right?

Answer 7

Kinda yes, i want to determine the image of the function, but without looking at the graph for the function.

Answer 8

Well, now we have all the features, positive after x =1, slowly increasing to 2 and just test the values between 0 and 1

Answer 9

the image is the range correct?

Answer 10

the image should be (from a translation from Danish to English) the interval of all values f(x) can take

Answer 11

then that is what is commonly refered to as range over here.

Answer 12

what are you endpoints defined as? when x=0, and the limit as x approaches infinity?

Answer 13

It's Brit/Euro for range (range c

Answer 14

Oh. I only know it as "værdimængde" and sense my math book is not in English then it was hard to look up the right word. But sorry for that misunderstanding

Answer 15

range causes confusion because some use it to mean codomain....

Answer 16

at x=0; f(0) = -1/2
as x approaches inf; f(x) approaches 2

this is only one step, do you see the rational in those values?

Answer 17

Sure do, but i can only see that solution and technique is valid as long there is no solution to the equation f ' (x) = 0?

Answer 18

ah, so you do know about deriavtives ..

Answer 19

Sure do. then I could esay find maxima and minima ^^

Answer 20

\[f(x) = (2x^2-x-1)* (x^2+3x+2)^{-1}\]
\[f'(x) = (4x-1) (x^2+3x+2)^{-1}-(2x^2-x-1)(2x+3)(x^2+3x+2)^{-2}\]
\[f'(x) = \frac{(4x-1) (x^2+3x+2)-(2x^2-x-1)(2x+3)}{(x^2+3x+2)^{2}}\]
\[f'(x) = \frac{7x^2+10x+1}{(x^2+3x+2)^{2}}\]
\[7x^2+10x+1=0\]
\[x=\frac{\sqrt{72}-10}{14}\]since sqrt(72) is less than 10, the max/min is not within the interval is it

and hopefully i mathed it up correctly

Answer 21

do we have any values that make the denominator go zero? and if so, can they be removed?

Answer 22

there is no offending zeros in the stated domain; so this is simply a function that goes from -1/2 to 2 along the stated domain

Answer 23

Why you take the derivative?

Answer 24

the derivative tells us the max and min spots of the function

Answer 25

You don't need that, you can just look at the function, no?

Answer 26

if there are no max or min within the stated domain, the endpoints are considered

Answer 27

without a graph, just looking at a function is not as simple

Answer 28

analysing the function is needed

Answer 29

I agree normally, but the domain is restricted, that makes it much easier..

Answer 30

yes, but without knowing if a min or max exists within the restricted portion ...

Answer 31

It's always increasing (by inspection) so only 0 to 1 to worry about....

Answer 32

how do you determine that it is always increasing (by inspection of what?)

Answer 33

You have 2x^2/x^2......

Answer 34

that is its horizontal asymptote; it limiting value as x approaches infinity; horizontal asymptotes can be crossed intermediately

Answer 35

unless there is a thrm that i cant recall ... id have to analyze the rest of the function

Answer 36

Yes but you can see that after x is 1, the function is always positive....

Answer 37

Well that would be with the derivative becuase we then can tell if there are positive and negative vaules (btw thanks so far amistre64 i sure understand your point, i just thought there was... well more behind it then just useing differential calculus)

Answer 38

spose a function graphs like this; after a certain value of x, the function is always positive right? but it would still have a max within a stated interval