**find all stationary points of f(xy) = xe^-x (y^2-y)**

x^2=-4 --> No Solutions Hope this is right, a long time since I touched this yeah thats right, since to find stationary points you need to do dy/dx=0 anyway .... C4 Differentiation - Stationary points PhysicsAndMathsTutor.com . 1. A curve has equation . x2 + 2xy – 3y2 + 16 = 0. Find the coordinates of the points on the curve where . x y d d = 0. (Total 7 marks) 2. A O x y. The diagram above shows a graph of . y = x v sin . x, 0 < x < ?. The maximum point on the curve is . A. (a) Show that the . x-coordinate of the point A satisfies the equation 2

**how do you solve 0=e^x (for x)? Yahoo Answers**

If you take the ugly mess of a 4th degree polynomial function that you get for a first derivative, you will find that you will be successful performing synthetic division three successive times with a divisor of 2. That will leave you with a simple linear equation the solution of which is the remaining stationary point.... If you take the ugly mess of a 4th degree polynomial function that you get for a first derivative, you will find that you will be successful performing synthetic division three successive times with a divisor of 2. That will leave you with a simple linear equation the solution of which is the remaining stationary point.

**How do you find the critical points for xe^x? Socratic**

Solve this equation to find the x-value of the stationary point. 3x(x ? 4) = 0 x = 0 or x = 4: Step 4. Substitute this value into the original function to find the y-value of the stationary point. When x = 0, y = 0 ? 0 = 0. When x = 4, y = 64 ? 96 = -32. Coordinates of the turning points are (0, 0) and (4, -32) Step 5. Examine the gradient on either side of the stationary point to find... For a function y = f(x) of a single variable, a stationary (or critical) point is a point at which dy/dx = 0; for a function u = f(x 1, x 2, , x n) of n variables it is a point at which In the case of a function y = f(x) of a single variable, a stationary point corresponds to a point on the curve at which the tangent to the curve is horizontal.

**find all stationary points of f(xy) = xe^-x (y^2-y)**

For a function y = f(x) of a single variable, a stationary (or critical) point is a point at which dy/dx = 0; for a function u = f(x 1, x 2, , x n) of n variables it is a point at which In the case of a function y = f(x) of a single variable, a stationary point corresponds to a point on the curve at which the tangent to the curve is horizontal.... x^2=-4 --> No Solutions Hope this is right, a long time since I touched this yeah thats right, since to find stationary points you need to do dy/dx=0 anyway .

## How To Find Stationary Point Of Y Xe X

### How to find the stationary point of $f(xy)=\\sin x \\sin

- Finding intercepts stationary points and point of
- How to find the stationary point of $f(xy)=\\sin x \\sin
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## How To Find Stationary Point Of Y Xe X

### Examples of Stationary Points Here are a few examples of stationary points, i.e. finding stationary points and the types of curves. Example 1 : Find the stationary point for the curve y = x …

- looks like, i got 1 stationary point (havent got my work on me so i can tell you peeps, i think i got it to be 2 or soemthing, either 2,1/2 or -2). Thanks Thanks 0
- Find the stationary points of the following functions: (a) g (y) = y-1 y 2-y +1 (b) g (x) = xe 2 x 2. Recall that if the demand function is given by Q = f ( P ) then the elasticity of demand is defined as E = P Q dQ dP For the following demand functions find the price level for which the elas- …
- looks like, i got 1 stationary point (havent got my work on me so i can tell you peeps, i think i got it to be 2 or soemthing, either 2,1/2 or -2). Thanks Thanks 0
- 22/10/2011 · Best Answer: f(x) = x*e^(-x^2) f '(x) = -2x^2*e^(-x^2) + e^(-x^2) Stationary pts occur where f '(x) = 0. So -2x^2 + 1= 0 x = +/- v2/2 So the coordinates are: (v2

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