Find an equation for the horizontal tangent to the curve y=x-3sqrt x
Can anyone please explain to me how to do this?I am kind of confused with the calculations.
THANKS A LOT!!!
The horizontal tangent is a tangent line where the slope dy/dx is zero.
Solve dy/dx = 1 - (3/2)/sqrtx = 0
sqrt x = 2/3.
Compute the value of y there.
y = (2/3) - 3sqrt(2/3) = 2/3 - sqrt6 = -1.7828
That is the tangent line equation, if I did the numbers correctly. Check my work.