How can I find the range of the function ?

In summary, to find the range of a function, one method is to compute its inverse and use its domain as the range of the original function. However, this can be problematic as the steps may not be reversible, resulting in extraneous solutions and a non-one-to-one inverse. Another method is to use calculus and find the critical points of the function, which can be easier and more accurate in some cases.
  • #1
Nanas
39
0
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



[itex]f(x) = \frac{e^x - e^{-x}}{2}[/itex]

After computing the inverse we have,

[itex]f^{-1} (x) = ln(x+\sqrt{x^2 +1})[/itex]

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

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

and we also note that [itex] f^{-1} [/itex] 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

[itex]g(x) = \sqrt{x+2} -1[/itex]

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
[itex]g^{-1} (x) = x^2 +2x -1 [/itex]
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
[itex]g (g^{-1} (x) )= x [/itex] for all x in the domain of [itex]g^{-1}[/itex]
we know that,
[itex] g (g^{-1} (x) )= x [/itex] if [itex] x\geq -1 [/itex]

[itex] g (g^{-1} (x) )= -x [/itex] if [itex] x < -1 [/itex]

then ,
[itex]g (g^{-1} (x) )= x [/itex] if and only if [itex] x\geq -1 [/itex]
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

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

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
 
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  • #2


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: [itex]f(x) = \frac{x^2 -1}{x-1}[/itex].

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

[itex]f(x) = \frac{e^x - e^{-x}}{2}[/itex]

[itex]f’(x) = \frac{e^x + e^{-x}}{2}[/itex], which has no zeros, so the function is monotonic, so you only have to consider infinity and negative infinity
 
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  • #3


Yes I know Calculus But I began learning it from only a week and I understand what you mean . Thanks for helping me
 
  • #4


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

Do you know what critical points are?
 
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  • #5
for your question and for allowing me to think about it.

I cannot comment on the mathematical methods you have described as it is not my area of expertise. However, I can suggest some general approaches to finding the range of a function.

1. Graph the function: Plotting the function on a graph can give you a visual representation of its range. The range of the function will be the set of all possible y-values on the graph.

2. Analyze the function: Look at the behavior of the function and determine its minimum and maximum values. This can give you an idea of the range of the function.

3. Use algebraic techniques: For some functions, you can use algebraic techniques to determine the range. For example, for a quadratic function, you can use the vertex formula to find the minimum or maximum value, which will give you the range.

4. Use calculus: Calculus can also be used to find the range of a function. By taking the derivative of the function and finding its critical points, you can determine where the function is increasing or decreasing and therefore, its range.

In conclusion, finding the range of a function may require different approaches depending on the function itself. It is important to have a good understanding of the behavior of the function and use appropriate mathematical techniques to determine its range.
 

1. What is the range of a function?

The range of a function is the set of all possible output values, or y-values, that the function can produce for any given input value, or x-value.

2. How can I determine the range of a function?

To determine the range of a function, you can graph the function and look at the y-values that correspond to different x-values. Alternatively, you can algebraically manipulate the function to solve for the output values.

3. Can a function have multiple ranges?

No, a function can only have one range. This is because for each input value, there can only be one corresponding output value.

4. What is the difference between the domain and range of a function?

The domain of a function refers to the set of all possible input values, or x-values, while the range refers to the set of all possible output values, or y-values.

5. Are there any shortcuts or tricks for finding the range of a function?

There are various techniques and shortcuts for finding the range of a function, such as using symmetry or identifying any asymptotes. However, the most reliable way is to graph the function or use algebraic methods.

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