Find the two stationary points on the curve when y = 2x/x^2 + 1
A stationary point occurs when the derivative of the curve is 0.
Find the derivative, y'. Solve for x when y'=0. Evaluate y at these values of x.
To find the stationary points on the curve, we need to find the values of x at which the derivative of y with respect to x is equal to zero. Let's start by finding the derivative of y.
Step 1: Find the derivative of y:
To find the derivative of y, we will use the quotient rule. The quotient rule states that if y = u/v, then the derivative of y with respect to x (dy/dx) is given by:
dy/dx = (v * du/dx - u * dv/dx) / v^2
For our function y = 2x/(x^2 + 1), let's label u = 2x and v = x^2 + 1.
du/dx = 2 (because the derivative of 2x with respect to x is 2)
dv/dx = 2x (because the derivative of x^2 + 1 with respect to x is 2x)
Applying the quotient rule, we obtain:
dy/dx = [(x^2 + 1) * 2 - 2x * 2x] / (x^2 + 1)^2
Step 2: Set dy/dx equal to zero and solve for x:
Setting dy/dx = 0, we have:
0 = [(x^2 + 1) * 2 - 2x * 2x] / (x^2 + 1)^2
Multiplying both sides by (x^2 + 1)^2, we get:
0 = 2(x^2 + 1) - 4x^2
Simplifying the equation further, we have:
0 = 2x^2 + 2 - 4x^2
Combining like terms, we obtain:
0 = -2x^2 + 2
Rearranging the equation, we get:
2x^2 = 2
Dividing both sides by 2, we have:
x^2 = 1
Taking the square root of both sides, we obtain two possible solutions for x:
x = 1 or x = -1
Step 3: Find the corresponding y-values:
To find the corresponding y-values for the stationary points, substitute the values of x into the original equation y = 2x/(x^2 + 1).
For x = 1:
y = 2(1)/(1^2 + 1) = 2/2 = 1
For x = -1:
y = 2(-1)/((-1)^2 + 1) = -2/2 = -1
Therefore, the two stationary points on the curve are (1, 1) and (-1, -1).
To find the stationary points on a curve, you need to find the points where the gradient (or slope) of the curve is zero. In other words, you need to find the points where the derivative of the function with respect to x is zero.
Let's start by finding the derivative of the function y = 2x/(x^2 + 1). To do this, we can use the quotient rule:
The derivative of y with respect to x, written as dy/dx, can be found by applying the quotient rule:
dy/dx = (2(x^2 + 1) - 2x(2x))/ (x^2 + 1)^2
dy/dx = (2x^2 + 2 - 4x^2)/(x^2 + 1)^2
dy/dx = (-2x^2 + 2)/(x^2 + 1)^2
Now, let's find the stationary points by setting the derivative equal to zero and solving for x:
(-2x^2 + 2)/(x^2 + 1)^2 = 0
Since the numerator of the fraction is equal to zero, we have:
-2x^2 + 2 = 0
Simplifying this equation, we get:
2x^2 = 2
x^2 = 1
x = ±1
So, we have found two possible values for x: x = 1 and x = -1. These are the x-coordinates of the stationary points on the curve.
To find the corresponding y-coordinates, we substitute these x-values back into the original equation:
For x = 1:
y = 2x/(x^2 + 1) = 2/(1^2 + 1) = 2/2 = 1
For x = -1:
y = 2x/(x^2 + 1) = 2/((-1)^2 + 1) = 2/2 = 1
Thus, the two stationary points on the curve are (1, 1) and (-1, 1).