A function whose upper bound would have attained its upper limit at a bound.
For example, f(x) = x - a whose domain is a < x < b
The upper bound is upper bound is b - a but, because x < b, the bound is never actually attained.
No. Not all functions are continuous. For example, the function f(x) = 1/x is undefined at x = 0.
There are continuous functions, for example f(t) = e^{t^2}, for which the integral defining the Laplace transform does not converge for any value of the Laplace variable s. So you could say that this continuous function does not have a Laplace transform.
Wherever a function is differentiable, it must also be continuous. The opposite is not true, however. For example, the absolute value function, f(x) =|x|, is not differentiable at x=0 even though it is continuous everywhere.
"Removable discontinuity" means the function is not defined at that point (it has a "hole"), but by changing the function definition at that single point, defining it to be certain value, it becomes continuous. "Irremovable discontinuity" means the function makes a sudden jump at that point. There are infinitely many functions like that; for example, you can set the function to be: f(x) is undefined at x = -2 f(x) = 0 for x < 2 (except for x = -2) f(x) = 1 for x > 2
Subject + had + been + present participle For example, I had been singing.
How about f(x) = floor(x)? (On, say, [0,1].) It's monotone and therefore of bounded variation, but is not Lipschitz continuous (or even continuous).
Yes. A well-known example is the function defined as: f(x) = * 1, if x is rational * 0, if x is irrational Since this function has infinitely many discontinuities in any interval (it is discontinuous in any point), it doesn't fulfill the conditions for a Riemann-integrable function. Please note that this function IS Lebesgue-integrable. Its Lebesgue-integral over the interval [0, 1], or in fact over any finite interval, is zero.
An infinite sum of continuous functions does not have to be continuous. For example, you should be able to construct a Fourier series that converges to a discontinuous function.
Let f(x)=abs(x) , absolute value of x defined on the interval [5,5] f(x)= |x| , -5 ≤ x ≤ 5 Then, f(x) is continuous on [-5,5], but not differentiable at x=0 (that is not differentiable on (-5,5)). Therefore, the Mean Value Theorem does not hold.
Yes.If you find 2 relative minima and the function is continuous, there must be exactly one point between these minima with the highest value in that interval. This point is a relative maxima.Think of temperature for example (it is continuous).
A continuous linear decreasing function is a line that goes on forever and has a negative slope (is downhill from left to right). For example, the line y = -x is a continuous linear decreasing function.
A function may have a finite number of discontinuities and still be integrable according to Riemann (i.e., the Riemann integral exists); it may even have a countable infinite number of discontinuities and still be integrable according to Lebesgue. Any function with a finite amount of discontinuities (that satisfies other requirements, such as being bounded) can serve as an example; an example of a specific function would be the function defined as: f(x) = 1, for x < 10 f(x) = 2, otherwise
The probability distribution function (pdf) is defined over a domain which contains at least one interval in which the pdf is positive for all values. Usually the domain is either the whole of the real numbers or the positive real numbers, but it can be a finite interval: for example, the uniform continuous distribution. Also, trivially, the pdf is always non-negative, the integral of the pdf, over the whole real line, equals 1.
No. Not all functions are continuous. For example, the function f(x) = 1/x is undefined at x = 0.
Apparently f(x)=sqrt(x)But I'm not sure why. That's what I'm looking for now :)
No. If the variable is continuous, for example, height or mass of something, or time interval, then the set of possible outcomes is infinite.
That means that the functions is made up of different functions - for example, one function for one interval, and another function for a different interval. Such a function is still a legal function - it meets all the requirements of the definition of a "function". However, in the general case, you can't write it as "y = (some expression)", using a single expression at the right.