How do you distinguish between a linear, quadratic, and cubic function?
601313
Yes, there is - called inflexion points. A linear equation has none, all points lie on a straight line; a quadratic has one, while a cubic has 2 inflexion points.\n\nInflexion point is a point where the curve basically changes direction. Thus, a quadratic equation, if you plot it on an X/Y graph, will rise (or fall) to a highest (or lowest) point, and then will fall (or rise).\n\nThus, a quadratic graph looks like an U (or an inverted U), while a cubic will look like an S on its side.\n\nMultiple regression technique is a way to see the effect of several variables on an outcome that you are interested in. By this method, you can find out how much is the result influenced by one or more of the variables.\n\nThere are several sites where this is explained in more details, if you google it. One such site is http://www.statsoft.com/textbook/stmulreg.html\n\nHope this helps
356793
A linear function can refer to two slightly different concepts. In geometery and elementary algebra a linear function is a first degree polynomial mathematical function of the form:\n\nf(x) = m x + c \nwhere m and c are constants.\n\nThe problem with this geometric definition is that functions of the above form - despite their names - do not necessarily satisfy the conditions of a linear map. Therefore, some people refer to functions of the above form as affine functions. If and only if a function is of the above form with c equal to zero, the function satisfies the properties of a linear map, preserving scalar multiplication and vector addition for all points in its domain.\n\nLinear functions always have as domain the set of all real numbers and a range of all real numbers. By the geometric definition, the derivative of linear functions in terms of their independent variable, x, is always the constant m.\n\nLinear functions (according to the geometric definition) can also be written in the form:\n\ny = m x + c \nand plotted on an x,y graph. It forms a straight line, as the name implies.\n\nThe constant m is often called the slope or gradient while c is the y-intercept, which gives the point of intersection between the graph of the function and the y-axis.
612525
We can have a simple linear function such as F(X)= X times a Constant C. This type of function is used everyday when we convert "one thing" to "another". One currency to another (i.e. exchange rate); metric and a myriad of other conversions. How about taxes like the sales tax. You pay so much per gallon of gas; so much per unit of power used for electricity etc.\nHowever we can also have a function F(X1,X2,X3...Xn)= C which can be quite complicated and which is a function of many inputs X1,X2,X3....Xn. Such a function could be used in computer modelling for weather and economic forecasts, navigation, traffic control etc. hope that helps somewhat
411263
AC or DC\n\nany phase differences\n\ndependent sources\n\nlinear or non linear elements\n\napplicability of theorems\n\nbilateral or not\n\nhomogeneity
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
693558
On a graph with x and y-axes , if you plot out the values of y calculated from an equation involving x squared you draw out a parabola.  A quadratic equation is one in the form y = a.x-squared + b.x + c \n a can be positive or negative (not zero or you don't involve x-squared). If negative, the parabola is upside down (open end downward).\nb and c can be positive negative or zero.\nWhen you are asked to "solve" a quadratic, you are finding where the parabola cuts the x-axis, that is to say the values of x where the equation calculates out to y = 0
666213
the answer is linear function:\nf(n)=2n-1
1122728
in standard ntation we have:\n\nA=2\nB=sqrt[3]\nC=1\n\ntan2t = B/(C-A) = -sqrt[3]\nHence 2t=-60 deg\nt=-30 deg\n\nThe standard notation for a quadratic curve is:\nAx^2+Bxy+Cy^2+Dx+Ey+F=0\nA, B, C determine the shape of the curve.\nThey are affected by rotation.\nchanging D, E, F only changes the position of the curve in plane by displacement.
911422
The discriminant is b^2 - 4ac. If it looks familiar, it's in the quadratic formula.\n\nThe quadratic is in the form ax^2 + bx + c. Plug in your values for a, b, and c.\n\nIf your result is:\n....positive, there are two real solutions\n...zero, there is one real solution (that is repeated)\n...negative, there are two imaginary solutions.\n\nHope that helped.
949364
since the second derivative function is a constant, it follows that the function is quadratic of the form ax^2 + bx + c = f(x)\nf(1) = 1 so a+b+c=1\nf'(x) = 2ax+b => f'(1) = 2a+b = 3 => b=3-2a\nf''(x) = 2a = 2 so a=1\n=> b=3-2(1) = 1\n=> (1) + (1) + c = 1 => c=-1\n\nf(x) = x^2 + x - 1\n\ni assume you made a mistake -- you said "show that f''(x)=x^2 + x - 1"\ni took that to mean f(x) not f''(x), for you just stated that f''(x)=2 for all x on the interval
714493
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
895128
If by 'convert' to a linear function you mean approximate by a linear function, then a possible solution is given by the Taylor's expansion of the exponential function.\n\nConsider a point x0, then the linear approximation of y around x0 is given by:\n\ny(x) = e^(x0+a) + e^(x0+a) * (x+a-x0) ,\n\nwhich is linear in x.  Obviously this is only an approximation as the Taylor expansion of the exponential function will involve a very large number of higher order (i.e. quadratic, cubic etc) terms in the right hand side of the equation.
791991
Given a(x^2) + bx + c as the quadratic the c determines the vertical shift of the graph. If c = 0 you get a parabola whose vertex touches (0,0). If c > 0 the graph shifts up to have a y-intercept of (0,c). If c <0 you get the same result but (0,c) falls below the x-axis.
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.