What is an example story for a cubic funtion?
606307
Suppose you have a rectangular box.\nThe length is 3 more than the height.  The width is twice as long as the height.\nWhat is the volume of the box?
554760
I do not know about F.But I know (x).Does she have something like(3,4)(3,5)?\n\nx_|y_ <-- or does she have that?\n3, 4\n\ny=2f(x)\n4=2f(3)                          \n\n      ^\n      That's what she is supposed to do.\n      To solve the function.\n     But still i am sorry about not knowing F.\n    (Those # is example)
152511
Which function, Subsitution, Linear COmbination, Graph?
838861
A linear example:\n\nThe demand for some types of products declines linearly with price.  The variables are Price on the x-axis and Quantity demaded on the y-axis.  The line has a negative slope\n\nAnother linear example:\n\nThe speed of your car is approximately a linear function of the force you exert on the pedal.  If you put pedal force on the x-axis and speed on the y-axis, the line will have a positive slope.
694927
What are you trying to do?  You have just defined f(x).  You have not said what to do with it.
819945
1.  the idea here is to take the limit for both pieces and show that they match.  This is relatively straight forward.  The limit as x approaches 1 from the left of x cubed is simply "1"     \nThe limit as x approaches 1 from the right of 3x-2 is 3(1)-2 also equals 1.  Since these two match, we have proven that "f is continuous at 1"\n\n2. The fact that f is continuous at 1 is INsufficient to imply that is is differentiable there\n\n3. the derivative from the right is simple 3 since the derivative of 3x-2 is 3.  The derivative from the left of 1 is also 3 since:\n\nf(x) = x^3 implies that f'(x) = 3x^2 and f'(1) = 3(1)^2 = 3 as well.  Since the derivative from the right equals teh one from teh left, the function is differntiable at x =1
979434
What you've got there is a function of a function, so you need to use the old chain rule on it. (Remember also, that if F(x) = kf(x), where k is a constant, then F'(x) = kf'(x).)\n(I take it x^3 means "x cubed".)\nSo, given that the derivative of ln y is 1/y, this makes the derivateive of 2 ln x^3 = (2/x^3) *3 x^2, and you can tidy up a bit and get (6x^2)/x^3 which cancels down to 6/x.
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for example\na/b\n---- = ad/bc\nc/d\n \n\na/b + c/d = (ad +cb)/ad\n\na/b -c/d =( ad- bc) /ad\n\n(a/b)(c/d)= ac/bd\n\nx
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
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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
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Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
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Lets see what values we can have for f(x). Suppose f(x) = c has some solution for x.\nThen (x^2 - 1) = c(x+2), so x^2 - cx - (2c+1) = 0. For that to have a solution, we need the discriminant, (b^2-4ac, though note thats a different c!) to be at least 0. So we need c^2+4(2c+1) >= 0.\nSo c^2 + 8c + 1 >= 0.\nNow, we can write that as (c+4)^2 - 15 >= 0, so (c+4)^2 >= 15.\nThus c+4 > sqrt(15) or c+4 < -sqrt(15).\nSo c > sqrt(15)-4 or c <-sqrt(15)-4.\n\nSo f(x) > sqrt(15)-4 or f(x) < -sqrt(15) - 4.\n\n(Note you could have done the last step solving the 'c' inequality with the quadratic formula; they're both the same, really.)\n\nNote that it is a much better idea to leave the inequality until as late as possible. Note that in the above idea of gp4rts, although it may be made to work, you *can't* do the following:\n(x^2-1)/(x+2) >= a\nTherefore\n(x^2-1) >= a(x+2).\nIf you multiply by a negative, you have to switch the inequality. Thus you would have to divide everything into two cases - if x+2>0, or x+2<0, and it gets very messy.
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Homework.\n\nShow it first for intervals. Then extend to simple functions. Then use the fact thats imple functions are dense in L1.
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Try the plot-o-matic, it helps in just this type of instance.\n\nGood luck with your story :-)
868371
can you give examples (such as the witches equation)
956017
well, to differentiate a the given funtion with respect to                (1-4x^2)^1/2, we assume (1-4x^2)^1/2=y. and the function becomes sin^-1(4y) or 1/sin4y. we have to differentiate this with respect to y.\nso, the function, which is cosec4y, on differentiation gives the result as -cosec4y*cot4y*4\nsubstituting the value of y we get the final result in terms of x as\n-4cosec((1-4x^2)^1/2)*cot((1-4x^2)^1/2)\nthe long string in the cot is (1-4x^2)^1/2 ie., y
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The easiest function of x are:\n\nf(x) = constant\nf(x) = a*x, where a is a constant : 1st order poly\nf(x) = b*x^2 where b is a constant 2nd order\nf(x) = c*x^3 where c is a constant: 3rd order\n...\nf(x) = z*x^n, where z is a constant, n is a positive integer, nth order polynomial\n\na general polynomial is a sum of these.  They are convenient to work with for lots of reasons.  That's why they are so important.