How can I find the range of the function ?

[Nanas]

how can I find the range of the function ??

I don't know a method to get the range of any function without the graph. But all I can say that for some functions I can get its range by first computing the inverse of the function , and this is the problem itself because in sometimes when computing the inverse the steps are not reversible,and thus the domain of the produced function is not the the same as the rage of the function, because we have extraneous solutions,and the produce function is not one-to-one,then we should know the range of the function in order to restrict the domain of the produced function.I will put examples



f(x) = \frac{e^x - e^{-x}}{2}

After computing the inverse we have,

f^{-1} (x) = ln(x+\sqrt{x^2 +1})

since our steps are reversible then these functions are invertible in order to check that directly from the definition after checking we have,

f(f^{-1} (x) ) = x for all x in the domain of f^{-1}
f^{-1} (f (x)) = x for all x in the domain of f (x)

and we also note that f^{-1} is one to one.
and thus we can know the range of f(x) directly from the domain of its inverse ,that,s because the domain of the inverse is the range of the function as if we tried to compute the inverse of the inverse we will have the the function back with the same domain and thus its rage equal the domain.(is my reasoning and concept is correct??)

Another example

g(x) = \sqrt{x+2} -1

I know it is easy to know the range of this function by applying transformations to sqrt(x)
But let's assume you don't know the range of g(x), then computing inverse
g^{-1} (x) = x^2 +2x -1
here is the problem we know when we computed the inverse we are possibly suppose extraneous solutions for y,and we note also that the inverse of g is not one to one, and thus can't be returned back to g ,so we must restrict the domain of g inverse , in order to be invertible, during my course in Algebra , we used to restrict this domain according to the range of g which is [-1 , infinity) but I assumed that we don't know this range.here I will go through the definition.we must have
g (g^{-1} (x) )= x for all x in the domain of g^{-1}
we know that,
g (g^{-1} (x) )= x if x\geq -1

g (g^{-1} (x) )= -x if x < -1

then ,
g (g^{-1} (x) )= x if and only if x\geq -1
then we restrict the domain of g inverse (-1, infinty) and because -1 is the x- coordinate of the vertex of g inverse so we have checked that g inverse is one-to-one
we also check, after checking

g(g^{-1} (x) ) = x for all x in the domain of g^{-1}
g^{-1} (g (x)) = x for all x in the domain of g (x)

so arguing as before we have the range of g (x) is [-1 , infinity )

so is these methods true, is my concept is correct ??

and thanks

 

[JonF]

Your general idea looks good. But you have to be careful, sometimes algebraic manipulation can alter the domain, which you have to take into account. Consider: f(x) = \frac{x^2 -1}{x-1}.

Do you know any calculus? Since the function is continuous you can just find the critical points which is often easier

f(x) = \frac{e^x - e^{-x}}{2}

f’(x) = \frac{e^x + e^{-x}}{2}, which has no zeros, so the function is monotonic, so you only have to consider infinity and negative infinity

 

[Nanas]

Yes I know Calculus But I began learning it from only a week and I understand what you mean . Thanks for helping me

 

[JonF]

Yes I know Calculus But I began learning it from only a week

Do you know what critical points are?