What is the difference between "y as a function of x" and "x as a function of y"?
Definition: A function is a set of ordered pairs of numbers (u,v) such that to each value of the first variable (u) there corresponds a unique value of the second variable (v). Given a set of ordered pairs, y being a function of x does not imply that x is a function of y. Consider the set {(0,0), (1,0)} f(x) = 0 for x = 0, 1. This satisfys the definition of y as a fuction of x, but not the definition of x is a function of y since the for the value y = 0 there are 2 corresponding x values.
You have to remember we like precision in mathematics, what that question is really asking is do you remember the definition of a function and application of that definition.Check out this phrasing on math is fun, it might make more sense.Definition of FunctionAs people mentioned you have to check if there is going to be one y for one x.if you have a math degree, the problem is easy, looking over the equation a you can notice that it is y that is squared so more then likely you will have two values for x for one y, and so one example is going to be sufficient to disprove the definition.part b is more complected and you would have to prove that it satisfies the definitionof a function, one way to do that is pick a random x, value represented by another variable and substitute it in, and then justify that expression you get represents one single value with continuity, increasing, decreasing. on the interval.here are the steps if you don’t have the math degree and don’t know operations as well it will work every time.re-write it in terms of y=…., meaning solve for yif it looks like you have two equations then it is a relationship -case aif you need to prove that one has a function for case b, you would point out that if you substitute any value you will get some value in Reals.
Functions; f inverse of 8/(x-3).?
If you want to find the inverse function of, for example f(x) their composition of x equals x. So the argument of f is the value of the f^(-1) and vice-versa. So do this. Solve this equation by x. 8/(x-3)=y (value of the function f(x) name y, a new variable, and determinate how much x equals in the terms of y.) y(x-3)=8 yx - 3y =8 yx = 8 + 3y x = (8 + 3y)/y So f^(-1)(y) = (8 + 3y)/y = 8/y+3 or f^(-1)(x) = 8/x+3. Now compose these two functions, and you will have to get x as a value. f(f^(-1)(x)) = f ((8/x+3) = 8/(8/x+3-3)=8/(8/x)=8x/8=x. Find the second one on your own, using the same algorithm. Tips: Define new variable again, y, and solve this equation by x: 12/x-4 = y. Once you find x in the terms of y, that value of x (once again, expressed in the terms of y) is the value of g^(-1)(y). And then, just change the letter to g^(-1)(x) and the value becomes expressed in the terms of x. (Last step is just changing the letter, it doesn't change the result, you can use any letter for any variable, x,y,z... are just common)
A function's injectivity and surjectivity depends upon the domain and co-domain. A can be injective for a given domain but can be not-injective . As the domain is [0, inf) and the co-domain is RThe above function is one-one(injective) but not onto (surjective).However if the co-domain too have been [0,inf) this function would have been surjective. And if the domain have too have been R ,|x| would not have been injective.An injective function is a function that preserves distinctness , it never maps distinct elements of its domain to the same element of its co-domain.as |x|= x in [0, inf)f(x)=x in the given domainLets assume for two different x f(x) is equalsf(x1)=f(x2)=> x1=x2Hence f(x1) = f(x2) only if x1=x2Moreover if you draw the graph it will be a straight line with slope 45 degrees. And if we draw a line parallel to x-axis it intersects the plotted line only at one place which means there are no two values in domain i.e. x for same value of y.Hence the function is one-one in the given domain.A function is said to be surjective if every element in Y (co-domain) has a corresponding element in X (domain)Now as the co-domain is R but the f(x) can take only positive values therefore it is not onto. As for all negative values in the co-domain there is no x in the domain mapping these values.As f(x) =x in [0,inf) As x>=0 Therefore f(x) >=0 Hence in the given Domain and for the given Co-domain the function f(x)= |x| is injective but not surjective.
Assume that f is a differentiable function of u, v and w. Let u = x - y , v = y - z and?
1) By the Chain Rule, ∂f/∂x = ∂f/∂u ∂u/∂x + ∂f/∂v ∂v/∂x + ∂f/∂w ∂w/∂x .......= ∂f/∂u * 1 + ∂f/∂v * 0 + ∂f/∂w * -1 .......= ∂f/∂u - ∂f/∂w. ∂f/∂y = ∂f/∂u ∂u/∂y + ∂f/∂v ∂v/∂y + ∂f/∂w ∂w/∂y .......= ∂f/∂u * -1 + ∂f/∂v * 1 + ∂f/∂w * 0 .......= -∂f/∂u + ∂f/∂v. ∂f/∂z = ∂f/∂u ∂u/∂z + ∂f/∂v ∂v/∂z + ∂f/∂w ∂w/∂z .......= ∂f/∂u * 0 + ∂f/∂v * -1 + ∂f/∂w * 1 .......= -∂f/∂v + ∂f/∂w. So, ∂f/∂x + ∂f/∂y + ∂f/∂z = (∂f/∂u - ∂f/∂w) + (-∂f/∂u + ∂f/∂v) + (-∂f/∂v + ∂f/∂w) = 0. ------------------------- 2) Use implicit differentiation. Taking derivatives of both sides with respect to x (holding y fixed): yz + xy ∂z/∂x = -sin(x + y + z) * (1 + ∂z/∂x) Solving for ∂z/∂x: yz + xy ∂z/∂x = -sin(x + y + z) - sin(x + y + z) ∂z/∂x ==> [sin(x + y + z) + xy] ∂z/∂x = -sin(x + y + z) - yz ==> ∂z/∂x = [-sin(x + y + z) - yz] / [sin(x + y + z) + xy]. Similarly, taking derivatives of both sides with respect to y (holding x fixed): xz + xy ∂z/∂y = -sin(x + y + z) * (1 + ∂z/∂y) Solving for ∂z/∂x: xz + xy ∂z/∂y = -sin(x + y + z) - sin(x + y + z) ∂z/∂y ==> [sin(x + y + z) + xy] ∂z/∂y = -sin(x + y + z) - xz ==> ∂z/∂y = [-sin(x + y + z) - xz] / [sin(x + y + z) + xy]. I hope this helps!
Find the directional derivative of f at P in the direction of v?
Please help i cant seem to figure out c and e? Given the function f (x,y) = x/y the point P =(1,-3) and the vector v = 3 i+ 2 j. (a) Find the gradient of f. f = 1/y i + -(x/y^2) j (b) Find the gradient of f at the point P. Gradient f(P) = -1/3 i + -1/9 j (c) Find the directional derivative of f at P in the direction of v. Du_f =? (d) Find the maximum rate of change of f at P. Answer = .3513641845 (e) Find the (unit) direction vector in which the maximum rate of change occurs at P. u = ? i + ? j
Find the linearization L(x,y) of the function at the given point????
Find the partial derivatives: ƒx(x,y) = ∂/∂x [ x·√(y) ] = √(y) · ∂/∂x [ x ] = √(y)·(1) = √(y) ƒy(x,y) = ∂/∂x [ x·√(y) ] = x · ∂/∂x [ √(y) ] = x/[ 2·√(y) ] ---- Evaluate the partial derivatives at the given point: ƒx(2,9) = √(9) = 3 ƒy(2,9) = 2/[ 2·√(9) ] = 2/( 2·3 ) = 2/6 = 1/3 Evaluate the original function at the given point: f(2,9) = 2·√(9) = 2·3 = 6 ---- Linearization formula for two dimesions: L(x,y) = f(x‚, y‚) + ƒx(x‚, y‚)(x - x‚) + ƒy(x‚, y)(y - y‚) Applying the above: f(2, 9) + ƒx(2, 9)(x - 2) + ƒy(2, 9)(y - 9) = (6) + (3)(x - 2) + (1/3)(y - 9) = 6 + 3·x - 6 + (1/3)·y - 3 = 3·x + (1/3)·y - 3 ---- Answer: L(x, y) = 3·x + ⅓·y - 3
An equation is something which puts things into perspective by providing an equal sign between these things. For example: 5+5 = 10 and yes your x+y = 10 is also an equation. When are things not an equation? Lets say you have something which looks like this e^x = 0. This is not true for any real number. So we say this equation does not hold true. Any expression which has one variable dependent on the value one or more other variables is a function of these other variables. Consider your equation "x + y = 10". y = x-10, So, any value of y depends on any value of x. So we say y is a function of x. We can also write it as f(x) = x - 10, where f(x) is a representation for the function. You could also have function like y = x+z+a. Here, f(x, z, a) = x+z+a is a representation for the function involving three variables.
Find all the second partial derivatives. v=(xy)/(x-y)?
g=dv/dx and h=dv/dy then we want to find dg/dx dg/dy dh/dx dh/dy so first lets find g and h dv/dx=[y(x-y)-xy]/(x-y)^2 dv/dx=-y^2/(x-y)^2 g=-y^2/(x-y)^2 dv/dy=[x(x-y)+xy]/(x-y)^2 dv/dy=x^2/(x-y)^2 h=x^2/(x-y)^2 dg/dx=2y^2/(x-y)^3 dg/dy=[-2y*(x-y)^2-2y^2(x-y)]/(x-y)^4 dg/dy=(x-y)[-2y(x-y)-2y^2]/(x-y)^4 dg/dy=[-2xy+2y^2-2y^2]/(x-y)^3 dg/dy=-2xy/(x-y)^3 dh/dx=[2x(x-y)^2-2x^2(x-y)]/(x-y)^4 dh/dx=2(x-y)[x(x-y)-x^2]/(x-y)^4 dh/dx=2[x^2-xy-x^2]/(x-y)^3 dh/dx=-2xy/(x-y)^3 dh/dy=x^2*-1*2(x-y)^-3 dh/dy=-2x^2/(x-y)^3
Which is the independent and dependent variable? X and Y?
Usually we call x the independent variable (the abscissa) and y the dependent variable (the ordinate). This would mean "y depends on x," like if y is set to f(x) = x², x is free to be whatever, and f(x) comes as a result. But this is simply a convention, and anyone could define the variables differently if they wanted...