Find critical points for y'=2/3x^3-4x
To find the critical points of the function y' = (2/3)x^3 - 4x, we need to first find the derivative, and then set it equal to zero.
1. Take the derivative of the function:
To find the derivative of y', we can use the power rule for differentiation. According to the power rule, the derivative of x^n is nx^(n-1).
dy/dx = d/dx (2/3)x^3 - 4x
= (2/3)(d/dx)(x^3) - (d/dx)(4x)
= (2/3)(3x^2) - 4
= 2x^2 - 4
2. Set the derivative equal to zero:
Now, we set the derivative equal to zero and solve for x to find the critical points:
2x^2 - 4 = 0
Adding 4 to both sides, we get:
2x^2 = 4
Dividing both sides by 2, we get:
x^2 = 2
Taking the square root of both sides, we obtain two solutions:
x = √2 and x = -√2
So, the critical points are x = √2 and x = -√2.