hectictar

avatar
Usernamehectictar
Score7324
Stats
Questions 8
Answers 2370

 #1
avatar+7324 
+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. smiley

hectictar Jul 29, 2017