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I am stuck on graphing

$y= (x-3) \sqrt{x}$

I am pretty sure that the domain is all positive numbers including zero. The y intercept is 0, x is 0 and 3.

There are no asymptotes or symmetry.

Finding the interval of increase or decrease I take the derivative which will give me

$\sqrt{x} + \frac{x-3}{2\sqrt{x}}$

Finding zeroes for this I subtract $\sqrt{x}$ and then multiply by the denominator

$2x = x - 3$

This gives me 3 as a zero, but this isn't correct according to the book so I am stuck. I am not sure what is wrong.

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1 Answer 1

up vote 3 down vote accepted

Your solution of $y'=0$ is incorrect. $$\begin{align*} \sqrt{x}+\frac{x-3}{2\,\sqrt{x}}&=0\\ \frac{x-3}{2\,\sqrt{x}}&=-\sqrt{x}\\ x-3&=-2\,x\\ x&=? \end{align*}$$

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No, it is correct because I am left with 2x-x = 3 which gives me x(2-1) = 3 which gives me 3 as a zero. –  user138246 Apr 1 '12 at 17:55
    
@Jordan It's $2x+x=3$, which gives $x=1$. –  David Mitra Apr 1 '12 at 17:59
    
I am incredibly bad at this. –  user138246 Apr 1 '12 at 18:10

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