y = Root(x) * (3-x)2
Find the x-coordinates of all points on y = Root(x) * (3-x)2 where the tangent is horizontal.
The first derivative GIVES the grandient of the tangent at any specific or general point. It is very important that you understand and remember this!
If the tangent is horizontal then the gradient of the tangent is 0.
So yes, you are being asked where dy/dx=0
\( y = \sqrt x * (3-x)^2\\ y = x ^{0.5}* (3-x)^2\\ \frac{dy}{dx}=0.5x^{-0.5}(3-x)^2\quad+\quad x^{0.5}*2(3-x)*-1\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}*\frac{2x^{0.5}}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad \frac{4(3-x)x}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2-4(3-x)x}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)[(3-x)-4x]}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)(3-5x)}{2x^{0.5}}\quad\\\)
This will equal zero when 3-x=0 and when 3-5x=0
That is when x=3 and when x= 0.6
Substitute these points back inot the original equation to find the vaues of y.
\(When\;\; x=3\\ y=\sqrt3(0)=0\\ When\;\; x=0.6\\ y=\sqrt{0.6}(3-0.6)^2\\ y=\frac{\sqrt{3}}{\sqrt5}\frac{12^2}{5^2}\\ y=\frac{144\sqrt{3}}{25\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}\\ y=\frac{144\sqrt{15}}{125}\\~\\ \text{The tangent to the curve will be horizontal}\\ \text{ at the points} (3,\frac{3}{5})\;\;and\;\;(\frac{3}{5},\frac{144\sqrt{15}}{125})\)
You should check my working. I have not done so there could be errors.