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. 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.
You need to find all values of x for which the first derivative of the function is greater than zero. Example: f(x)=x^2+5x+6 f'(x)=2x+5 = 0 x=-5/2 Consider intervals (minus infinity, -5/2) and (-5/2, infinity). Choose any number in each interval and examine the sign of f'(x), that is 2x+5. If it is positive, the function increases on that interval. If it is negative, the function decreases on that interval. In this case, the function increases on (minus infinity, -5/2) and decreases on (-5/2, infinity).
the zeros of a function is/are the values of the variables in the function that makes/make the function zero. for example: In f(x) = x2 -7x + 10, the zeros of the function are 2 and 5 because these will make the function zero.
There are many families of functions or function types that have both increasing and decreasing intervals. One example is the parabolic functions (and functions of even powers), such as f(x)=x^2 or f(x)=x^4. Namely, f(x) = x^n, where n is an element of even natural numbers. If we let f(x) = x^2, then f'(x)=2x, which is < 0 (i.e. f(x) is decreasing) when x<0, and f'(x) > 0 (i.e. f(x) is increasing), when x > 0. Another example are trigonometric functions, such as f(x) = sin(x). Finding the derivative (i.e. f'(x) = cos(x)) and critical points will show this.
A one-to-one graph must pass both the horizontal and the vertical line test. That means that no x-value can have two y-values and no y-value can have two x-values. An example of a one-to-one function is a line. Things like parabolas and the graph of an absolute function cannot be one-to-one.
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.
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 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.
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.
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.
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).
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.
what is the example of conclusion
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.