what are the coordinates of the points at which the slope of the function with the equation y = x^4 - 8x^2 is equal to zero ?
y' = 4x^3 -16x = 4x(x^2-1)
0 at x = 0
0 at x = 1
0 at x =-1
so at
( 0, 0 )
( 1, -7
(-1, -7)
y' = 4 x^3 - 16 x = 4x (x^2 - 4)
zero at x = 0, 2 , -2
so
(0, 0)
(-2,-16)
(+2,-16)
To find the coordinates of the points where the slope of the function y = x^4 - 8x^2 is equal to zero, we need to find the derivative of the function and determine where it equals zero.
Step 1: Find the derivative of the function.
The derivative of x^n, where n is a constant, can be found using the power rule. The power rule states that the derivative of x^n is n * x^(n-1).
In this case, the derivative of y = x^4 - 8x^2 can be found by taking the derivative of each term separately:
dy/dx = 4x^3 - 16x
Step 2: Set the derivative equal to zero and solve for x.
To find the values of x where the slope is zero, we set dy/dx = 0 and solve for x:
4x^3 - 16x = 0
Step 3: Factor out common terms.
To solve this equation, we can factor out the common term x:
x(4x^2 - 16) = 0
Step 4: Solve for x.
Setting each factor equal to zero gives us two possible values of x:
x = 0 (from x = 0)
4x^2 - 16 = 0
4x^2 = 16
x^2 = 4
x = ±2
So, the coordinates of the points where the slope of the function y = x^4 - 8x^2 is equal to zero are (0, 0) and (2, 0) and (-2, 0).