Given F X Y =10xy^2 0

What is the difference between f(x,y) =0; f(xy) =0; f(x/y) =0? What exactly do the variables inside the brackets indicate?

The function f(x,y) is a bivariate function; that means f is defined over two independent variables x and y.For easier imagination, consider z=f(x,y) in the x,y,z coordinate system.Let's define z=x^2 + y^2 = f(x,y)For z=0, x^2 + y^2 = 0 represents the x=0 and y=0 in z=0 plane, that is it represents the originFor z=1, x^2 + y^2 = 1 represents a circle in z=1 planeFor z=2, x^2 + y^2=2 represents a circle in z=2 plane and so on...So, as z moves from 0 to ∞, z=x^2 + y^2 represents a paraboloid with the z-axis as the axis of the paraboloid. That means, f(x,y) is a paraboloid x^2 + y^2 in which x and y can take values independent of each other.If we say, f(x,y)=0 then it represents only one point on the parabola:Origin.You can take more such examples, where z= x^2 - y^2 or z=x + y^2 etc and imagine what the result will be when z=0. Like, if we take z=x^2 - y^2, then z=(x+y)(x-y)=0 represents a pair of straight lines.Note that, I have introduced z to represent f(x,y) for easier understanding and imagination.Now, if we take f(xy)=0, then it basically is a function in terms of 'xy', that is a function which takes values as product of x and y and not individually. If we introduce, u=xy, then f(xy)=0 can be given by f(u)=0, where u is the product of x and y. It still is a single variable function where the domain is given by 'u' where u = xy for all real x and y such that f(u) = 0The same goes with f(x/y), where it is a function in terms of 'x/y' . If we introduce, v=x/y, then f(x/y)=0 can be given by f(v)=0

Given the function f(x) =x^2-3bx+(c+2), determine the values of b and c such that f(1) =0 and f'(3) =0?

This can be solved by following the steps given below:[math]f(x) = x^2 - 3bx + (c+2)[/math]If [math]f(1)=0[/math][math]\implies f(1) = 1^2 - 3b(1) + (c+2)[/math][math]\implies 0 = 1-3b+c+2)[/math][math]\implies 0 = 3-3b+c[/math][math]\implies 3b-c = 3 ----------------------------------------[/math](1)If [math]f(3)=0[/math][math]\implies f(3) = 3^2 - 3b(3) + (c+2)[/math][math]\implies 0 = 9-9b+c+2)[/math][math]\implies 0 = 11-9b+c[/math][math]\implies 9b-c = 11 ----------------------------------------[/math](2)Subtracting (1) from (2).[math]\implies 6b = 8[/math][math]\implies b = \dfrac{8}{6}[/math][math]\implies b = \dfrac{4}{3}[/math]Putting value of b in (1):[math]\implies 3\dfrac{4}{3} - c = 3[/math][math]\implies 4 - c = 3[/math][math]\implies c = 4 - 3[/math][math]\implies c = 1[/math]Therefore the values of b and c are [math]\dfrac{4}{3}[/math] and 1 respectively.

What is the difference between y=-f(x) and y=f(-x)?

f(-x) means that you replace every ‘x’ by -x.-f(x) means that you change + or - of f(x).It is the same function, if the function only has x with odd exponents like x, x^3, x^5, x^(-7) etc.However, if you have anyhing else, f(-x) is not -f(x).For example, f(x)= x^2.f(-x) would be (-x)^2 = x^2, as the minus x positivize each other.-f(x) would be -(x^2), so -x^2, as you simply multiple the function by (-1).If the functions includes numbers independet from x, like +3 or -4, f(-x) doesn’t change them and -f(x)= makes them -3 or +4.You’ll notice the difference primarly on exponential functions.Let f(x) be 4^x.f(-x) would be equal to 4^(-x)-f(x) simply equals to -4^x.So in this case f(-x) decreases towards 0 and -f(x) decreases to minus infinity.Tl;dr -f(x) means multiplieng everything with minus one and f(-x) replacing every x with -x.

The equation of a curve is y=x^2. What is the equation of the tangent at point (3,1)?

I'll assume you want a tangent to the curve that goes through that point. The equation of a line through that point is [math]y-1=m(x-3)[/math]  where m is the slope of the parabola at the point where the line in question is tangent to it. Let's call that point's x value t. Then [math]m=2t[/math] because the derivative of [math] y=x^2[/math] is [math]2x[/math]. Thus [math] y=2t(x-3)[/math] Since this line is tangent at t, it goes through the point [math] (t, t^2)[/math]. We plug in that point to get [math]t^2-1=2t(t-3)[/math]. This can be written [math] 0=t^2-6t+1[/math]. Use the quadratic formula to find the two solutions [math] 3\pm 2\sqrt{2}[/math]. Thus, the two lines that work are [math]y=(6\pm 4\sqrt{2})(x-3)+1[/math].See the graph: Desmos Graphing Calculator

How can I maximize [math]1000x+300y+700z+100xy+200xz+50yz+10xyz[/math] with the constraint [math]x+y+z=50[/math] and [math]x, y, z \ge 0[/math]?

For finding a local maximum (or minumum), which may or may not be the global maximum (or minimum), all partial derivatives for all variables must be zero. A partial derivative for a variable is a derivative where all other variables are considered to be constants.Let[math]f = 1000x+300y+700z+100xy+200xz+50yz+10xyz[/math]You need to solve the following set of equations:[math]\displaystyle\frac{\partial f}{\partial x} = 1000 + 100y + 200z + 10yz = 0[/math][math]\displaystyle\frac{\partial f}{\partial y} = 300+100x + 50z + 10xz = 0[/math][math]\displaystyle\frac{\partial f}{\partial z} = 700 + 200x + 50y + 10xy = 0[/math]Oh wait, there are additional constraints? Then this is a constrained optimization problem. It is even a non-linear one, because [math]f[/math], the cost function to be maximized, is non-linear (whereas the two constraints are linear in the variables). Therefore, the Simplex algorithm cannot be used.However, you may be lucky: solving the problem without the constraints might lead to a solution which even respects the constraints.

What are good ways to solve the equation f(f(x)) = x? How many solutions exist?

Well, if we already have some function [math]f_0(x)[/math] which satisfies the equation [math]f_0(f_0(x))=x[/math], then it follows that for any invertible function [math]g(x)[/math] then [math]g^{-1}(f_0(g(x)))[/math] also satisfies it!It's simple enough to prove this by simply plugging our new solution into the functional equation:[math]g^{-1}(f_0(g(g^{-1}(f_0(g(x))))))[/math]Cancel the inverses:[math]g^{-1}(f_0(f_0(g(x))))[/math]Because [math]f_0[/math] is already a solution, we know that [math]f_0(f_0(x))[/math] will just cancel out:[math]g^{-1}(g(x)) = x[/math]Therefore [math]F(x)=g^{-1}(f_0(g(x))) [/math]is a solution for [math]f_0[/math] a solution and [math]g[/math] any invertible function.What does that mean? Well, we already know that [math]k-x[/math] and [math]k/x[/math] are solutions for any k (in the latter for non-zero k). So from these solutions we can generate infinitely many others!Example: let [math]f_0[/math] be [math]1-x[/math] and [math]g[/math] be [math]x^3[/math]. Then our new solution would be:[math]F(x)=\sqrt[3]{1-x^3}[/math]Which, when graphed, is clearly symmetric with the line [math]y=x[/math]However, I'm not entirely sure if this captures every possible function. I do know that it captures every continuous function with an invertible derivative, however.EDIT: To clarify my last paragraph, look at the original equation:[math]f(f(x))=x[/math]Apply implicit differentiation and the chain rule[math]f'(f(x)) \cdot f'(x) = 1[/math]With a little algebra we get:[math]f(x)=f'^{-1}(\frac{1}{f'(x)})[/math]Which is exactly what I found before, with [math]f_0[/math] being [math]1/x [/math] and [math]g [/math]the derivative of the function!