- #1
Calculuser
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I have a question about domain and range.What are the domain and range of [itex]cos(e^{-×})[/itex] and [itex]\frac{\left|2×-1\right|}{sin(\frac{1}{2}\pi×-\pi)}[/itex] ??
Thanks..
Thanks..
Calculuser said:The first question's solution:
[itex]f(x)=cos(e^{-x})[/itex]
If we take [itex]lim_{x\rightarrow+∞}cos(e^{-x})\approx1[/itex]. However if we take [itex]lim_{x\rightarrow-∞}cos(e^{-x})[/itex] the function takes values on closed interval -1 and 1 [itex]\left[-1,1\right][/itex]. That's why we can easily say that [itex]f:ℝ\rightarrow\left[-1,1\right][/itex].
Second one:
[itex]g(x)=\left|2x-1\right|/sin(\frac{\pi}{2}x-\pi)[/itex]
For this function denominator can not be zero [itex]sin(\frac{\pi}{2}x-\pi)\neq0[/itex] So that, we have to except the such x values [itex]x=\left\{x:x=2k,k\in Z\right\}[/itex] that make it zero. That was easy part. The hard one is to find the set of range of this g(x) function. I'm actually stuck in that. Help for [itex]g:ℝ-\left\{x:x=2k,k\in Z\right\}\rightarrow?[/itex] ??
Would you mind confirming the form of the second? It seems unnecessarily complicated, but that would make more sense if it were [itex]\frac{\left|2x-1\right|}{sin(\frac{1}{2}\pi-\pi{x})}[/itex]Calculuser said:I have a question about domain and range.What are the domain and range of [itex]cos(e^{-×})[/itex] and [itex]\frac{\left|2×-1\right|}{sin(\frac{1}{2}\pi×-\pi)}[/itex] ??
jbriggs444 said:Limits? Why would you care about limits? The question is about the function and its values on its defined domain, not about the values of some hypothetical extension onto some hypothetical extended domain that includes plus or minus infinity.
jbriggs444 said:What is an inverse of f(x) = cos(e−x)?
It's a fairly easy inverse to define -- at least symbolically.
The inverse of cos is arccos.
The inverse of e^x is log
The inverse of - is -
Now you only have the problem of taking the log of a number that may not be strictly positive. But that can be dealt with.
HINT: The inverse of cos is multi-valued. You can pick which inverse you want.
Calculuser said:Why would not? Of course, with my self-study I realized that it's useful to find the range of a function is given. We can find the boundaries of a function through limit then make out what its range is.
I have no idea why you told me the inverse of the function? We don't need to use the inverse of the function to find the function's range and domain values.
jbriggs444 said:Taking a limit will not tell you what the boundaries of a function are. A limit, if it exists, is the value that the function f() approaches as its argument approaches x.
If the limit exists, it tells you only one thing about the domain of f() -- that the domain of f includes values that are arbitrarily close to (and not equal to) x.
If the limit exists, it tells you only one thing about the range of f() -- that the range of f includes values that are arbitrarily close to (and possibly equal to) the limit.
jbriggs444 said:If you are asking yourself whether a particular value of y is in the range of a function f() then one way to proceed is to try to explicitly find an x such that f(x) = y. Ideally you can come up with a formula that takes a value y and produces such an x. The range of f() is then equal to the domain of this formula.
Such a formula is an inverse of f().
There are other approaches -- such as the mean value theorem. That might be a way to formalize what you had in mind with the "limit" argument.
##f(x)=\frac{1}{x^2+1}##, f:R->RCalculuser said:What if I take [itex]\stackrel{lim}{x\rightarrow\mp∞}f(x)[/itex] then find the values which the function has boundaries between them. So I can say the limit of the function x goes to infinity either sides in other words that would determine the range of the function. Did you catch my point?
mfb said:[itex]f(x)=\frac{1}{x^{2}}, f:R->R
f(x)=\frac{sin(x)}{x^{2}}, f:R->R[/itex]
Both have limits of 0 for both sides. How does that help?
Calculuser said:What if I take [itex]\stackrel{lim}{x\rightarrow\mp∞}f(x)[/itex] then find the values which the function has boundaries between them. So I can say the limit of the function x goes to infinity either sides in other words that would determine the range of the function. Did you catch my point?
I know the meaning and rules of an inverse function. I meant it will not all the time be useful to try to find the domain and range of a function by solving from its inverse. If you try what you said on the functions I've posted here, you'll get what I meant. It just gets complicated.
The domain of a function is the set of all possible input values, or independent variables, for a given function. In other words, it is the set of values that can be plugged into the function and produce a valid output.
The range of a function is the set of all possible output values, or dependent variables, for a given function. In other words, it is the set of values that the function can produce.
To determine the domain of a function, you must consider the restrictions on the independent variable. These restrictions can include excluded values, such as dividing by zero or taking the square root of a negative number, as well as any other conditions that would make the function undefined.
To determine the range of a function, you must consider the possible output values for each input in the domain. This can be done by evaluating the function for different values of the independent variables or by using algebraic methods, such as finding the maximum and minimum values of the function.
The domain and range of a function are closely related. The values in the domain determine the possible input values for the function, which in turn determine the output values, or range, of the function. In other words, the domain and range are dependent on each other and are both important in understanding the behavior of a function.