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How do you prove a function is continuous on a closed set between 0 and 1?
Wiki User
∙ 14y ago
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.
Wiki User
∙ 14y ago
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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.
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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
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How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
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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.
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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.
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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?
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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
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