0
you need to show that there are no disconituities in the line between 0 and 1.
For instance all polynomials are continuous over the whole real line, however if you get a divide by zero in a fuction at any point then you will have an assymtote at that point, and as such a disconinutity.
What else can I help you with?
How do you prove if the given function is a constant function using Mean Value Theorem?
The Mean Value Theorem states that the function must be continuous and differentiable over the whole x-interval and there must be a point in the derivative where you plug in a number and get 0 out.(f'(c)=0). If a function is constant then the derivative of that function is 0 => any number you put in, you will get 0 out. Thus, using the MVT we deduced that the slope must be zero and since the f(x) is a constant function then the slope IS 0.
What does the horizontal line test prove?
I posted this question myself to be honest because i wasn't sure... but the horizontal line test was made to prove whether the function/graph was an one-to-one function
How would you prove algebraically that the function f(x) x-2 x 2 is one to one?
How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
What is the difference between show that and prove that?
there is no difference
How do you prove integers aren't closed under subtraction?
Take any two integers, and subtract one from another, you will have another integer. If there was a situation where you could show that this statement is not true, then that would prove your hypothesis, but I cannot think of any.
State and prove uniform limit theorem?
There exists an N such that for all n>N, for any x. Now let n>N, and consider the continuous function . Since it is continuous, there exists a such that if , then . Then so the function f(x) is continuous.
Let f be an odd function with antiderivative F. Prove that F is an even function. Note we do not assume that f is continuous or even integrable.?
An antiderivative, F, is normally defined as the indefinite integral of a function f. F is differentiable and its derivative is f.If you do not assume that f is continuous or even integrable, then your definition of antiderivative is required.
How would you prove that there are no closed contours in the contour plot of a harmonic function?
This is not exactly true as a constant function is harmonic and has closed contours as its contour plot (i.e. the entire plane is closed). However, any function that has closed contours can be shown to be the constant function. Here is how. If, say u, is a harmonic function which is constant on a contour which is closed, then the inside of that contour is a domain (simply connected set if that has any meaning to you). By the maximum and minimum principles respectively, the function u must attain both its max and min on the boundary i.e. the contour. This number is a constant and since the maximum is the same as the minimum we can conclude that the entire function is constant on the insides of the contour. From there we can extend this function to the entire plane by identity principle.
Prove that if the definite integral is continuous on the interval ab then it is integrable over the interval ab sorry that I couldn't type the brackets over ab because it doesn't allow?
why doesn't wiki allow punctuation??? Now prove that if the definite integral of f(x) dx is continuous on the interval [a,b] then it is integrable over [a,b]. Another answer: I suspect that the question should be: Prove that if f(x) is continuous on the interval [a,b] then the definite integral of f(x) dx over the interval [a,b] exists. The proof can be found in reasonable calculus texts. On the way you need to know that a function f(x) that is continuous on a closed interval [a,b] is uniformlycontinuous on that interval. Then you take partitions P of the interval [a,b] and look at the upper sum U[P] and lower sum L[P] of f with respect to the partition. Because the function is uniformly continuous on [a,b], you can find partitions P such that U[P] and L[P] are arbitrarily close together, and that in turn tells you that the (Riemann) integral of f over [a,b] exists. This is a somewhat advanced topic.
Prove that the boundary of a set is involved in that set only when this set is a closed set?
That is the definition of a closed set.
How do you prove if the given function is a constant function using Mean Value Theorem?
The Mean Value Theorem states that the function must be continuous and differentiable over the whole x-interval and there must be a point in the derivative where you plug in a number and get 0 out.(f'(c)=0). If a function is constant then the derivative of that function is 0 => any number you put in, you will get 0 out. Thus, using the MVT we deduced that the slope must be zero and since the f(x) is a constant function then the slope IS 0.
How do you proving one-to-one function?
The answer depends on what you wish to prove!
How would you prove algebraically that the following function is one to one?
How would you prove algebraically that the following function is one to one? f(x)= (x+3)^2 , x>= -3?
Are differentiable functions always continuous?
Differentiability implies continuity This is easy to prove using the limit of the difference quotient
What does the horizontal line test prove?
I posted this question myself to be honest because i wasn't sure... but the horizontal line test was made to prove whether the function/graph was an one-to-one function
How do you get a group home for boys closed?
You would need to prove there was no need for it or it was in breach of some statute
What is the purpose of the network security authentication function?
To require users to prove who they are
