For #f(x)=sin^2x# what is the equation of the tangent line at #x=(3pi)/2#?
1 Answer
Explanation:
graph{(y-(sinx)^2)=0 [-10, 10, -5, 5]}
Substitute in the x value to find the y value.
The coordinates of the point are
graph{((x-4.71225)^2+(y-1)^2-0.007)=0 [-10, 10, -5, 5]}
Find the equation for the derivative(slope)
graph{2sinxcosx [-10, 10, -5, 5]}
Substitute in the value of x to get a numeric value for the derivative(slope)
A slope of zero indicates a horizontal line.
Use the point slope formula to find the equation of the tangent line.
Now substitute in the values
graph{(y-(sinx)^2)(y-1)((x-4.71225)^2+(y-1)^2-0.007)=0 [-10, 10, -5, 5]}