VeeEight said:Yup you solve both
[itex]cos^2 x- 3 sin^2 x[/itex][itex]= cos^2 x- 3(1- cos^2 x)[/itex][itex]= cos^2 x- 3+ 3cos^2 x= 0[/itex]cruisx said:Ok so i got the values for the left but the right eqn is nto going well. I do not understand what to do after cos^2x - 3(1-cos^2x)
The equation for finding local max and min points for y = sinxcox^3x is to first take the derivative of the function, set it equal to 0, and then solve for x. The x-values obtained from this process will be the local max and min points.
To take the derivative of y = sinxcox^3x, you will need to use the product rule and the chain rule. The derivative will be 3cosxsin^2x + cos^3xsinx.
Yes, there are specific steps you should follow to solve for x in the derivative equation. First, factor out any common terms. Then, use the quadratic formula to solve for x. Finally, check your answer by plugging it back into the original equation.
Yes, you can use a graphing calculator to find local max and min points for y = sinxcox^3x. Simply enter the equation into the calculator and use the "zero" or "root" function to find the x-values where the derivative is equal to 0.
The local max and min points represent the highest and lowest points on the curve of the graph of y = sinxcox^3x, respectively. They are also known as turning points, where the direction of the curve changes from increasing to decreasing or vice versa.