# hectictar

+2
29
2
+8956
hectictar  May 29, 2020, 10:02 PM
+2
281
4
+8956
hectictar  Jan 22, 2020
+4
1
1064
3
+8956

hectictar  Sep 9, 2018
+4
1
1066
2
+8956

### Decomposing a Vector

hectictar  Apr 25, 2018
+11
1
1204
4
+8956

### Melody's Birthday!!!

off-topic
hectictar  Apr 15, 2018
+4
1
1064
2
+8956
hectictar  Mar 21, 2018
+7
2909
11
+8956
hectictar  Oct 2, 2017
+2
1
1607
3
+8956
hectictar  Sep 20, 2017
+1
1
1561
2
+8956

hectictar  Apr 25, 2017
+1
1
1559
4
+8956
hectictar  Mar 29, 2017
+2
1
1497
2
+8956
hectictar  Feb 22, 2017
#1
+8956
+3

Your derivative is correct...here are the steps...

$$\begin{array}\ y&=&x^{\frac12} \cdot(3-x)^2 \\ \frac{dy}{dx}&=&(\frac{d}{dx} x^{\frac12})(3-x)^2+(\,\frac{d}{dx}(3-x)^2\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(\,\frac{d}{dx}(3-x)\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(-1)(x^{\frac12}) \\ \frac{dy}{dx}&=&(3-x)(\quad(\frac12)(x^{-\frac12})(3-x)+(2)(-1)(x^{\frac12})\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(\quad(x^{-\frac12})(3-x)-4x^{\frac12}\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(\quad(3-x)-4x\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(3-5x) \end{array}$$

The above equation tells us what the slope of this curve is at any  x  .

Where the tangent line is horizontal, the  slope  of the curve  =  0 .

Where the tangent line is horizontal, the  dy/dx  of the curve  =  0  .

We want to know what values for  x  cause  dy/dx  to be  0  .

$$0\,=\,(3-x)(\frac12)(x^{-\frac12})(3-5x)$$

Set the first and fourth factors equal to zero and solve for  x .

0  =  3 - x                   0    =   3 - 5x

x  =  3                        5x  =   3

x    =   3/5

You're right that the minumum is a point where the tangent line is horizontal...but the maximum is also a point where the tangent line is horizontal.

Jul 29, 2017