What are the coordinates of the points at which the slope of a function with the equation y=x^4-8x^2 is equal to zero?
Please help me to understand how to approach this question.
y'=0=4x^3-16x= 4x(x+2)(x-2)
so what are the zeroes of this?
Is it (0,0), (2,0) and (-2,0)?
To find the coordinates of the points where the slope of a function is zero, we need to find the values of x for which the derivative of the function is equal to zero.
In this case, we're given the function y = x^4 - 8x^2. To find the derivative of this function, we use the power rule for derivatives.
The power rule states that the derivative of x^n is equal to n*x^(n-1).
Using the power rule, we calculate the derivative of y = x^4 - 8x^2:
dy/dx = d/dx(x^4) - d/dx(8x^2)
= 4*x^(4-1) - 8*2*x^(2-1)
= 4*x^3 - 16x
Now we set the derivative equal to zero:
4*x^3 - 16x = 0
Next, we factor out the common factor of 4x:
4x(x^2 - 4) = 0
Now we have two possibilities: either 4x = 0 or (x^2 - 4) = 0.
From 4x = 0, we find that x = 0.
From (x^2 - 4) = 0, we find two solutions:
1. x^2 - 4 = 0
x^2 = 4
x = ±2
So the coordinates of the points at which the slope of the function is zero are (0,0), (-2,0), and (2,0).