x^3-x, responses

Yemsrach Woubbie

The graph of y=x^3-x crosses the x axis three times, at -1, 0 & 1. From this we can understand that the graph has a local maxima in the second quadrant and a local minima in the forth quadrant. And this can be found as: y = x^3 - x Set dy/dx=0, dy/dx = 3x^2 - 1 = 0 This implies, x^2=1/3 Therefore; the local max is approximatly when x=0.577 and the local minima is approximatly when x=-0.557

Wanapon Techagaisiyavanit

dY/dX = X^3-X = (3X)^2 -1 0 = (3X)^2 -1 X = +,-(1/3)^(1/2) To find out the min and the max, take another derivative of (3X)^2 -1 dY/dX = 6X Plugging in both value: 6(1/3)^(1/2)=3.46 6(-1/3)^(1/2)= -3.46 Therefore, the negative value would be maxima and the positive value would be minima

Tasha Sakaguchi

If you take the derivative of the function you get 3x^2-1. If you set that to zero, you end up with a square root: +-(1/3)^1/2. So that it would be both plus and minus. Bcs it is a cube root, you end up with a shape that really does not have any global max or mins, but that thier local max and mins end up at the same point, where the derivative equals zero. so at both neg and pos (1/3)^1/2 giving you both max and min at the same time.

hilary moore

the local maxima and minima for x^3-x is found by finding the derivative: 3x^2=1, x^2=1/3, x=(1/3)^(1/2)or +/-.577. Therefore, the maxima is .577, and the minima is -.577.

Ashlea

y' = 3x^2 - 1. If y' = 0, the critical case... 0 = 3x^2 - 1, 1 = 3x^2, 1/3 = x^2. So the local maxima and minima in the critical cases are the positive and negative square roots of 1/3.

Alia Rahman-Khan

y = X^3-X 0

ingrid e. frau

derivative of y=x^3-x is 3x^2 - 1, set it to zero and find it in terms of x. it turns out to be the square root of 1/3, or .5773502 approximately. this is the minima. there is no maxima because x can always be made bigger.

collin hull

find derivative: 3x^2+1, set equal to 0: x= plus or minus the square root of (1/3). the local maximum is at -square root of (1/3) and the local minimum is at +square root of (1/3).

Catherine Chan

After pluging the function y=X^3-X into the calculator, the slope of the graph is first increased from teh left then decreased and increased on the right again. The graph contains both local maxima and minima. The local maxima is on the left-side of the graph, where ( -0.5773521, 0.38490018) The local minima is on the right of the graph where (0.57734878, -0.3849002)

Claire

By taking the derivative of the function y=x^3-x, you get y'=3x^2-1. In order to find the local maxima and minima, you set the derivative equal to zero and solve for x and you get x=.58 (the square root of 1/3). However when solving for x, you must keep in mind that because you have taken a square root, you have lost a root by ignoring the negative value of .58. By plugging in the x values to the original equation, you find that at -.58, y=.38 and at .58, y=-.38. So there is a local maximum at x=-.58 and a local minimum at x=.58.

Anuradha Tulachan

setting d(x^3-x)/dx=0 or 3x^2-1=0 or 3x^2=1 or x^2=1/3 x=+(1/3)^1/2,-(1/3)^1/2 therefore, local maxima is at -(1/3)^1/2 and local minima at +(1/3)^1/2

sarah fraser

1) the max of the problem is 1/radical 3 found this by solving for y' and then setting equal to zero and min = 1/2radical 3 by using same method of plugging back into problem

Youngshin Cho

when the function is y=x^3-x, the derivitive of this is y'=3x^2-1 so when the derivitive of the function and the origin function meets x=1/3+1/3^1/3 and 1/3-1/3^1/3 so the maxima is the 1/3+1/3^1/3 and the minima is the 1/3-1/3^1/3

stephanie

0=3x^2-1, 1=3x^2 divide three by both sides 1/3=x^2 then take the square root of both sides. and that gives you x = 1/square root of 3. then substitute 1/square root of 3 in for the x var. and you get -2/3square root 3. then substitute -1/the square root of 3 for x. and you get +2/3 square root 3

Bushra Husain

First find the derivative of the function: y'=3x^2-1 Set the derivative of the function equal to zero: 0=3x^2-1 Solve for x: 1=3x^2 square root of 1= square root of 3x square root of 1 divided by the square root of 3 Therefore, the maxima and the minima of the funtion are + and - the square root of 1 divided by the square root of 3.

Melanie J. LaFavre

There is one local minima and one local maxima for this problem. The are found in the two points that the derivative disaprears. y'=3x^2-1 soo y'=0 when x=+ or - the square root of 1/3. Which are the maxima and minima for the problem.

saima

The maxima of the function would be that amount that gives the largest result! To get the minima of the funtion one would have to input the smallest. To get the minima one can use all the numbers that are fractions - because a fraction raised to any power will make the number smaller, and thus the y coordinate smaller. For example .5^3 = .125 y = 0.125 - .5 = - 0.375 - and if one uses negative numbers then the answer will be the maxima. For example : -2^3 = 8 and then y = -2^3 - (-2) = 10

saima

Marium Khan

To get the minima one should plug in all fraction numbers - reason being that a fraction raised to any power will make the number smaller, and make the y coordinate smaller. For example .2^3 = .008 y = 0.008 - .2 = -.992 - To get the maxima number on the other hand, we should use negative numbers.

Anonymous

x^3-x d/(dx)(x^3-x)=3x^2-1 3x^2-1=0 3x^2=1 3x=+ or -1 x=+ or - 1/3

Anonymous

If I am reading this function right, as x is smallest at the point 0 but continues to rise on either side of the graph. X therefore hits it minima at 0 and has an undefined maxima, since it will continue to rise no matter what y is.

Mark

The derivative of this function is 3*X^2-1, which is zero at X=1/sqrt(3) and X=-1/sqrt(3). It seems to take a sketch of the graph to figure out which is a minimum and which is a maximum. The one at positive X is a minimum. The one at negative X is a maximum. That is suggested by evaluating the original function: when X=1/sqrt(3), X^3-X is -(2/3)/sqrt(3). At X=-1/sqrt(3), X^3-X is +(2/3)/sqrt(3), which is bigger.