which one of the following function is strictly bounded?
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A function f(x) is bounded if there are numbers m and M such that m≤f(x)≤M for all x . In other words, there are horizontal lines the graph of y=f(x) never gets above or below.
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We say that f(x) has a relative (or local) maximum at x=c if f(x)≤f(c) f ( x ) ≤ f ( c ) for every x in some open interval around x=c . We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on.
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In fact, affine functions are the only functions that are both convex and concave. Some quadratic functions: f(x) = xT Qx + cT x + d. – Convex if and only if Q ≽ 0. – Strictly convex if and only if Q ≻ 0.
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If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Given a function y=f(x) y = f ( x ) , the form y=f(bx) y = f ( b x ) results in a horizontal stretch or compression.
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We know that for f to be continuous, then ∀ϵ>>0, ∃δ>0 such that if x0 and x ∈S and |x−x0| < ϵ, |f(x)−f(x0)| < δ must hold. ... Thus, f(x)=x is uniformly continuous on R.
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Density theorems enable us to prove properties of Lp functions by proving them for functions in a dense subspace and then extending the result by continuity. For general measure spaces, the simple functions are dense in Lp. ... Suppose that (X, A,ν) is a measure space and 1 ≤ p ≤ ∞.
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But actually in those points the function tan(x) is not defined and hence they are not in the domain of tan function. So, it is ok to say of course tan(x) is continuous in its domain.
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2 Answers. This function is not one-to-one, since f(0)=f(−1)=0. The definition of a one-to-one function (also called injective sometimes) is that two different inputs are mapped to two different outputs. ... Whether or not the function is defined by a formula, piecewise, not continuous, or polynomial... it does not matter.
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1 Answer. This function is neither even nor odd, since neither f(−x)=f(x) nor f(−x)=−f(x) holds.
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['0≤PX(x)≤1 for all x;', '∑x∈RXPX(x)=1;', 'for any set A⊂RX,P(X∈A)=∑x∈APX(x).']
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Answer and Explanation: The natural log function, f(x) = ln(x) does not have a horizontal asymptote.
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O(1) and Θ(1) aren't necessarily the same if you are talking about functions over real numbers. ... This function is O(1) because for any n ≥ 1, f(n) ≤ 1.
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IN GENERAL , has NO INVERSE , IF IT IS NOT A ONE-TO-ONE FUNCTION.,because only such functions are invertible. ⑵ BUT if a cubic functionis is of the following form/can be converted to the following form, it is invertible : (i) f(x)=(ax+b)³+c, a≠0 , b,c∈|R, with its natural domain, x∈|R or a reduced domain.
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Most functions are neither. In this case, as a real-valued function with a subset of the real numbers as its domain, the domain can't include any non-positive values. It's already true that the function can't be even or odd (unless it's regarded as defined over an empty domain (!), when it is both).
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The set of all real numbers is the only interval that is unbounded at both ends; the empty set (the set containing no elements) is bounded. An interval that has only one real-number endpoint is said to be half-bounded, or more descriptively, left-bounded or right-bounded.
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A set is bounded if the distance between any two points in the set is less then some finite constant. A set in R^n is bounded if all of the points are contained within a disc of finite radius. ... A closed set is a bounded set that contains its boundary. A bounded set need not contain its boundary.
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The absolute value function is continuous at 0. Since |x| = -x for all x < 0, we find the following left hand limit. Since the two one-sided limits agree, the two sided limit exists and is equal to 0, and we find that the following is true. ... In fact, the absolute value function is continuous (on its entire domain R).