Which equations or inequalities is y a function of x? explain the reasoning.?
816555
all except 3)\nall other equations/inequalities describe y as an explicit function or dependence on the value of x.\nmoreover their is a linear relationship btw y and x. y can change its value depending on how the value of x changes because of the cnstraint i.e. presence of a constant(integer) in the eq/ineq\nfor eg 1. y = 1-x so whatever value u give to x, y takes 1-x value, sop is true for 4), 5). for 3) x-y =2, so even if x and y change their values independently their difference is limited by '2', that makes y dependent on x.\n3) describes x as a fn of y.
797160
Your function y=4*x^2*e^(5x) is defined for any x. Meanwhile x^2>=0 & e^(5x)>0, thus y>=0 for any x. Now y’=4*(2x+5x^2)*e^(5x), local extrema being at y’=0. Or x1=0, x2=-0.4\nHence and “above” x1=0 means local min & y(0)=0, so x2=-0.4 means local max, \nLim[for x->-infinity]=0 & Lim[for x->+infinity]=+infinity. Does it help you Xu?
700913
a function is written in  the form f(x) or g(X)  or anything relating to that form\nan equation has and equal sign\n\ntherefore this one is a linear equation
202630
people are not here to do your homework:\n\nThis is the definition of a function:\nA variable y is a function of x if a relation between x and y produces exactly one value of y for each value of x.\n\nread it carefully, the answer to your question is quite obvious
632327
Write f(x,y) as x^2*[y^2/(x^2+y^2)]. The second factor is always <=1, so f(x,y) goes to 0 as (x,y) goes to (0,0). The first answerer is WRONG. As for the second answerer, just knowing the limit is 0 along every line is not enough to conclude that the limit exists. It is possible for limits along lines to be 0, but to be otherwise along parabolas.\n\nIf f(0,0) is defined to be 0, then this function has partial derivatives at (0,0) that are both 0. In fact, f is differentiable at (0,0) in that case. The derivative is the 0 functional.
564677
x=2 y=1 so 1 = a*2^b\nx=6 y=9 so 9 = a*6^b = a*(2^b)*(3^b)\n\nbut a*(2^b)=1 from the first eq so\n\n9 = 3^b, this givez b=2\nfrom here, a=1/4\n\nthe function: y=(x^2)/4
795059
You came up with x > -1 OR x < 7.  The "or" is key.  It can be less than 7, or  greater than -1.  There's no real number that doesn't fit the solution.\n\nFor the second one, you came up with x >= 0.8 AND x >= 2.  Similarly, the "and" is the key here.  Anything greater than or equal to 2 is going to be greater than 0.8, fulfilling both conditions.  The same cannot be said if the numbers are reversed.\n\nTo sum it up, if the problem has an "or" condition, every solution must solve at least one inequality.  If the condition is "and", every solution must solve ALL inequalities.
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
26362
Part 1:\n\nWe can rewrite y^x as (e^ln(y))^x which is e^(xln(y))\n\n∂f/∂x f(x,y)= ∂f/∂x (x^y) + ∂f/∂x(y^x)\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)e^(xln(y))\n\n∂f/∂x f(x,y)= yx^(y-1) + ln(y)y^x\n\n\nPart 2: There is no simple function which is the integral of e^(x^2).
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
781968
I'm not sure if your question meant f(x)=2x(x-1)(x-4) or f(x)=x(x-1)(x-4)^2, so I will answer both.\n\n1) f(x)=2x(x-1)(x-4)\n(0, 1) U (4, ∞)\n\n2) f(x)=x(x-1)(x-4)^2\n(-∞, 0) U (1, 4) U (4, ∞)
379867
This is a typical function composition homework problem.\n\nExample:\nSuppose we have two functions of a variable x, \n\nf(x) = 1/x \n\nand\n \ng(x) = 1/x, \n\nthen \n\nf(g(x))\n= f(f(x)) \n= 1/(f(x)) \n= 1/(1/x)\n= x\n\nIt is given in this example that f(x) = g(x) = 1/x. Therefore, f(f(x)) = x. \n\nNow, f(f(x)) = f(x) \n\ncan be written \n\nx = 1/x.\n\nModifying this equation by cross multiplying give us x^2 = 1. Solving this equation by a method of your choice (factoring, quadratic formula, completing the square, square root property, etc.) gives x = 1 or x = -1.\n\nLet's check each of these answer. \nx = -1:\nf(-1) = 1/(-1) = -1 = x\n\nx = 1:\nf(1) = 1/(1) = 1 = x\n\nSo, there you have it: x = -1, 1.\n\nIf you have a graphing calculator, then graph y = x and  y = 1/x. Notice where the two graphs touch. That's right, at -1 and 1. \n\nDONE!
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
721733
if y is always increasing then y' is always >=0 (1) \ny'=3ax^2+2bx+c \ndelta': D=b^2-3ac (2)\nwe also have a>0 (3)\n(1),(2)&(3)--> D <=0-->b^2-3ac <=0-->b^2<=3ac (qed)\nNote: we have b^2 <=3ac, not b^2 < 3ac,for example : function y=(x-1)^3
790758
Quit thinking x and y and start thinking independent variable and dependent variable, or independent variable and function result.\n\n\nAll you care about is that the independent variable is along the abcissa, and the dependent variable is along the ordinate.\n\nThat said, just pretend that the y in your equation is x for the TI and the x in your equation is the y for the TI.
1099133
Let Y =  0 + x/(0 + x/( ...\nthen we can do a sneaky substitution and say that:\nY = 0 + x/Y\nSo, Y = x/Y\nx = Y^2\nY = +/- sqrt(x)\ng(x) = +/- sqrt(x)\nSince all the terms are positive, this is positive, so g(x) = sqrt(x).\nExcept for g(0), because then you would be dividing by 0, so g(0) is undefined.
916581
To write any equation in "function form", just solve for y (get y by itself on one side of the =), and replace y with f(x).\n\nI'll let you do that.  :)