What is the difference between functions and equations?
101134
A function always outputs a value. Suppose the function \nf(x) = 10x+20, what this means is that this function outputs different values for different input values of variable 'x'. For example:\nf(10) = 10X10+20 = 120\nf(20) = 10X20+20 = 220\ni.e. we substitute different values into x and get different outputs. \nThe important point is that every function outputs a value.\n\nComing to Equations. Equations define the relationship between one or more variables.\nSingle variable equation:\nx2 - 10x = 0\nMulti vairable equation:\n2x+3y = 10\nAn equation doesnt output a value. We can solve the equation and find out the values of the variables. The equation would be valid only for certain values of the variables.\n\nHope the difference is clear.
1084156
The answer to that is a line.  If you want a unique solution to a two-variable problem, you need two equations.
21931
It's like the difference between an exponent and a polynomial.\n\nDifferential equations are equations that can contain multiple derivatives (first derivative, second derivative, etc.), each with a coefficient. \n\nhttp://opera.answers.com/Differential%20Equations
437570
You're probably talking about the vertical line test.  A relation is a function iff you can draw a vertical line anywhere and intersect the function at just one point.\n\nSo, something like y = x is a function because a vertical line always hits just one point.  y^2 = x is not a function because if you draw a vertical line at x=4, it hits the graph at both (4,2) and (4,-2).
581102
You use derivatives to compute how values are changing. A good example is to compute velocities.\n\nIntegration is used to compute areas and volumes.
296652
Where's the function?
30225
These problems are talking about composite functions, which is where the results of one function are input into the second.  Usually it's written with a small circle in between (I'll use gof(x)) and read "g of f of x".  Remember that the function closest to the variable is the one that acts on it first.\n\nOne helpful tool to use when you can't figure out a problem is simplifying it to a problem you've done before.  For example, if f(x)= 3x and g(x)=x^2 ("x squared") then the composition gof(x) means take the results of f(x) and put them into g(x).  \nSo then gof(x) = (3x)^2 = 9x^2.\n\nYour problem is a bit more complicated, but we can use the same process:  \n\ngof(x) would be g(ax-1) so \ngof(x) = (ax-1)^2 + 3(ax-1) + 1\nUse the distributive property to expand this to\ngof(x) = a^2x^2 - 2ax + 1 + 3ax - 3 + 1\nand combine like terms to get\ngof(x) = a^2x^2 + ax - 1   *\n\nSo far, we've done everything just from the definitions of g(x) and f(x).  Now we use the last piece of information we were given: \nthat gof(x) = 4x^2 + ax + b  *\nWe compare the coefficients on like terms of these two polynomials (with the stars) to determine that since a^2 must equal 4, \na = 2 and b = -1 (from comparison of the constant terms).\n\n\nSee if you can use a similar process for the second problem!
305407
Yes! Big difference. Arithmetic has to do with simple operations of only numbers involving +, -, /, and x. (adding, subtracting, dividing, and multiplying). Mathematics encompasses EVERY kind of math, from arithmetic to calculus to linear algebra to the stuff they used for the physics to build the atomic bomb.
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All it is, is solving a function within a function.\nThe first thing you need to do is take the g(x) function and plug 3 in for X. The answer you get is 15  (if I read your equation correctly. I read it as: 2x squared - 3)\nSo now thats you know g(x) = 15 you plug 15 in for your X in the function f(x). You get: 49. That's you answer.\n\nNote: If I read your equation incorrectly then just follow these steps:\n1. Solve the Function g(x)\n2. Take that answer and plug it into the Function f(x)\n3. Viola! Your answer.\n4. Check out my homework help question, to see if you can help me. :P
867649
That is an equation, but y is a linear function of x.
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First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
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The domain for a function is all the "x" values that work in the functions.\n\nAn example is "y = x"\nIn this case you can plug anything in for "x" and get a "y"\n\nAnother example is "y = 1/x"\nHere, x cannot be zero (since you cannot have 1/0)\nTherefore the domain is all values except x = 0\n\nIn your equation, (1/x^2)  - 3x  + 2, the domain is all real numbers except x = 0.  If x were 0, you'd have (1/0) - 0 + 0 and you cannot evaluate that.
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To define an equation, you can use the syntax\nf := x -> (k*x*(1-x));\nWhich is creating a function defined on a variable x.\nThen, you can set the value of k by saying \nk := 2;\nIf you then say\nf(.1);\nit will give you the value of f(.1).  You can repeat this with the value you get from this.
1155111
xsquare=x+x+x...x terms\nby diff x\n2x=1+1+1...x terms\n2x=x\n2=1
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Definition:  A function is a one-to-one function if and only if each second element corresponds to one and only one first element. (each x and y value is used only once)\n\nUse the horizontal line test to determine if a function is a one-to-one function.\n\nIf ANY horizontal line intersects your original function in ONLY ONE location, your function will be a one-to-one function and its inverse will also be a function.\n\nTIP:\n\nGraph your function to see if it passes the horizontal line test.\nYou can do the horizontal line test with a pencil.  If the pencil crosses the graph of your function in more than one place (more than one point), then it will not be one-to-one.  If the pencil crosses the graph of your function at only one location, then it will be one-to-one.  \n\nTry it and get back to me.\n\nGuido
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Partial differential equation can be solved by two methods one is by Calculus type and other is by Numerical Methods.\n\nOne of the conventional way of programming the partial functions is used by the numerical methods the the equation is simpled to simple algebric equations and this equation is consider as the base for developing program. IN programming using looping, conditions assigning etc., the algrebic equation is more easier way.
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you mean f(x)=3x+2, for example?\n\nreplace f(x) with Y, graph and solve as a normal equation. Linear functions have infinite whole solutions, depending on what you want Y to be.\n\nOf course, I could be minunderstanding you completely. If so, sorry.