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From Wikipedia
In mathematics a linear inequality is an inequality which involves a linear function.
Linear inequalities in real numbers
Definitions
When two expressions are connected by 'greater than' or 'less than' sign,we get an inequation.
When operating in terms of real numbers, linear inequalities are the ones written in the forms
 f(x) < b \, or f(x) \leq b,
where f(x) is a linear functional in real numbers and b is a constant real number. Alternatively, these may be viewed as
 g(x) < 0 \, or g(x) \leq 0,
where g(x) is an affine function.
The above are commonly written out as
 a_0 + a_1 x_1 + a_2 x_2 + \cdots + a_n x_n < 0
or
 a_0 + a_1 x_1 + a_2 x_2 + \cdots + a_n x_n \leq 0
Sometimes they may be written out in the forms
 a_1 x_1 + a_2 x_2 + \cdots + a_n x_n < b
or
 a_1 x_1 + a_2 x_2 + \cdots + a_n x_n \leq b
Here x_1,\ x_2,...,x_n are called the unknowns, a_{0},\ a_{1},\ a_{2},...,\ a_{n} are called the coefficients, and b is the constant term.
A linear inequality looks exactly like a linear equation, with the inequality sign replacing the equality sign.
A system of linear inequalities is a set of linear inequalities in the same variables:
 \begin{alignat}{7}
a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; \leq \;&&& b_1 \\ a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; \leq \;&&& b_2 \\ \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; \leq \;&&& b_m \\ \end{alignat} Here x_1,\ x_2,...,x_n are the unknowns, a_{11},\ a_{12},...,\ a_{mn} are the coefficients of the system, and b_1,\ b_2,...,b_m are the constant terms.
This can be concisely written as the matrix inequality:
 Ax \leq b
where A is an m×n matrix, x is an n×1 column vector of variables, and b is an m×1 column vector of constants.
In the above systems both strict and nonstrict inequalities may be used.
Not all systems of linear inequalities have solutions.
Interpretations and applications
The set of solutions of a real linear inequality constitutes a halfspace of the ndimensional real space, one of the two defined by the corresponding linear equation.
The set of solutions of a system of linear inequalities corresponds to the intersection of the halfplanes defined by individual inequalities. It is a convex set, since the halfplanes are convex sets, and the intersection of a set of convex sets is also convex. In the nondegenerate cases this convex set is a convex polyhedron (possibly unbounded, e.g., a halfspace, a slab between two parallel halfspaces or a polyhedral cone). It may also be empty or a convex polyhedron of lower dimension confined to an affine subspace of the ndimensional space R^{n}.
Sets of linear inequalities (called constraints) are used in the definition of linear programming.
Linear inequalities in terms of other mathematical objects
When you graph a linear inequality, it will be on one side of a line. Also, when you mark points where the line crosses where the x and y axis cross each other you can make the rise over run, which will help you find slope. If slope is denoted by m and yintercept by b, you can find m = \frac {y_2  y_1}{x_2  x_1} and b = \frac {x_2y_1  x_1y_2} {x_2x_1}, so long as x_1 \neq x_2. Such line is described by the equation y = mx + b.
The above definition requires welldefined operations of addition, multiplication and comparison, therefore the notion of a linear inequality may be extended to ordered rings, in, particular, to ordered fields.
In mathematics, a linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants). The concept of linear combinations is central to linear algebra and related fields of mathematics. Most of this article deals with linear combinations in the context of a vector space over a field, with some generalizations given at the end of the article.
Definition
A linear combination is the sum of some set of ordered pairs (here, vectors), each ordered pair weighed (multiplied) by some real number. For example, if you had the ordered pairs regarding age and height, and you had the set {(5,30), (10,40), (50,70)}, an instance of linear combination would be 4*(5,30)+6*(10,40)+3*(50,70) which would equal (230,570).
Suppose that K is a field (a set of real numbers) and V is a vector space over K. As usual, we call elements of Vvectors and call elements of K scalars. If v_{1},...,v_{n} are vectors and a_{1},...,a_{n} are scalars, then the linear combination of those vectors with those scalars as coefficients is
 a_1 v_1 + a_2 v_2 + a_3 v_3 + \cdots + a_n v_n. \,
There is some ambiguity in the use of the term "linear combination" as to whether it refers to the expression or to its value. In most cases the value is emphasized, like in the assertion "the set of all linear combinations of v_{1},...,v_{n} always forms a subspace"; however one could also say "two different linear combinations can have the same value" in which case the expression must have been meant. The subtle difference between these uses is the essence of the notion of linear dependence: a family F of vectors is linearly independent precisely if any linear combination of the vectors in F (as value) is uniquely so (as expression). In any case, even when viewed as expressions, all that matters about a linear combination is the coefficient of each v_{i}; trivial modifications such as permuting the terms or adding terms with zero coefficient are not considered to give new linear combinations.
In a given situation, K and V may be specified explicitly, or they may be obvious from context. In that case, we often speak of a linear combination of the vectorsv_{1},...,v_{n}, with the coefficients unspecified (except that they must belong to K). Or, if S is a subset of V, we may speak of a linear combination of vectors in S, where both the coefficients and the vectors are unspecified, except that the vectors must belong to the set S (and the coefficients must belong to K). Finally, we may speak simply of a linear combination, where nothing is specified (except that the vectors must belong to V and the coefficients must belong to K); in this case one is probably referring to the expression, since every vector in V is certainly the value of some linear combination.
Note that by definition, a linear combination involves only finitely many vectors (except as described in Generalizations below). However, the set S that the vectors are taken from (if one is mentioned) can still be infinite; each individual linear combination will only involve finitely many vectors. Also, there is no reason that n cannot be zero; in that case, we declare by convention that the result of the linear combination is the zero vector in V.
Examples and counterexamples
Vectors
Let the field K be the set R of real numbers, and let the vector space V be the Euclidean spaceR^{3}. Consider the vectors e_{1} = (1,0,0), e_{2} = (0,1,0) and e_{3} = (0,0,1). Then any vector in R^{3} is a linear combination of e_{1}, e_{2} and e_{3}.
To see that this is so, take an arbitrary vector (a_{1},a_{2},a_{3}) in R^{3}, and write:
 ( a_1 , a_2 , a_3) = ( a_1 ,0,0) + (0, a_2 ,0) + (0,0, a_3) \,
 = a_1 (1,0,0) + a_2 (0,1,0) + a_3 (0,0,1) \,
 = a_1 e_1 + a_2 e_2 + a_3 e_3. \,
Functions
Let K be the set C of all complex numbers, and let V be the set C_{C}(R) of all continuousfunctions from the real lineR to the complex planeC. Consider the vectors (functions) f and g defined by f(t) := e^{it} and g(t) := e^{−it}. (Here, e is the base of the natural logarithm, about 2.71828..., and i is the imaginary unit, a square root of −1.) Some linear combinations of f and g are:
 \cos t = \begin{matrix}\frac12\end{matrix} e^{i t} + \begin{matrix}\frac12\end{matrix} e^{i t} \,
 2 \sin t = (i ) e^{i t} + ( i ) e^{i t}. \,
On the other hand, the constant function 3 is not a linear combination of f and g. To see this, suppose that 3 could be written as a linear combination of e^{it} and e^{−it}. This means that there would exist complex scalars a and b such that ae^{it} + be^{−it} = 3 for all real numbers t. Setting t = 0 and t = π gives the equations a + b = 3 and a + b = −3, and clearly this cannot happen. See From Yahoo Answers
Answers:probably the absolute value (which looks like a V) and the cube function (y = x^3)
Answers:You could have just added this in to the additional details.... Linear: f(x)=x this is a straight line (ie linear) it has on x intercept (0) these equations are found by graphic a constant speed on a time/distance graph. D is all real numbers and R is all real. Function is odd (ie symmetrical from (0,0) Absolute value: f(x)= x The general shape is a V, it as one x intercept at 0 (the parent function does but it can have up to two), the are seen when you are only looking for positive y answers (ie distance)..... D is any real numbers and R is anything greater than 0, the function is even (symmetrical from the y axis) Quadratic: f(x)=x^2 General shape is a U, parent only has one x intercept but can have 2, D is any real number and R is anything greater than 0.... Even Square Root: f(x)= sqrt(x) GS Steep closer to the origin but it flattens out (relative to the x axis) as you get farther away. One x intercept at 0. Domain is anything greater than 0 and range is any positive real number. Neither odd nor even Cubic: f(x)=x^3 GS Looks like an S that has been turned sideways and stretched (http://www.google.com/imgres?imgurl=http://library.thinkquest.org/2647/media/oddxxx.gif&imgrefurl=http://library.thinkquest.org/2647/algebra/ftevenodd.htm&usg=__BxjG9VWZU593Jp7jRgtxmGIqWoc=&h=317&w=365&sz=3&hl=en&start=0&sig2=c49qej0i_HjEhrwj6KVCVg&zoom=1&tbnid=fQob5GWIX1BdNM:&tbnh=126&tbnw=144&ei=xz6ZTPK4C4P88AaW89DvDw&prev=/images%3Fq%3Df(x)%253Dx%255E3%26um%3D1%26hl%3Den%26safe%3Doff%26biw%3D1440%26bih%3D673%26tbs%3Disch:1&um=1&itbs=1&iact=hc&vpx=128&vpy=74&dur=342&hovh=126&hovw=145&tx=90&ty=95&oei=xz6ZTPK4C4P88AaW89DvDw&esq=1&page=1&ndsp=31&ved=1t:429,r:0,s:0) One x intercept at 0,0 Domain is any real number and range is any real number... Function is odd Reciprocal: f(x)=1/x GShttp://www.mathsrevision.net/gcse/functions.php No x intercepts Domain is any real number besides 0 and range is any real number besides 0.... I hope THIS answers your question....
Answers:You must verify the three axioms of a norm. For any f and g in C[0,1] and any a in R, you should have:  a f  = a  f   f + g   f  +  g   f  = 0 if and only if f = 0 These are all fairly easy. The first is a computation:  a f  = max(  a f(x)  ) = max( a f(x) ) = a max( f(x) ) = a  f  The second is an application of the normal triangle inequality:  f + g  = max(  f(x) + g(x)  ) max(  f(x)  +  g(x)  ) max( f(x) ) + max( g(x) ) =  f  +  g  The third is also fairly obvious  if  f  = max( f(x) ) = 0, then f(x) 0 everywhere. That implies f=0. And from the definition, it is clear that 0 = 0. QED For part b, just notice that 3x+2 is an increasing function so  f  = f(1) = 5.
Answers:I don't believe so. They have different slopes, so there would have to be some sort of rotation or reflection.
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