Critical Number from Analying a Graph.

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Homework Statement



Find all critical numbers of f(x)= x\sqrt{2x+1}

Homework Equations



Derivative, Product Rule

The Attempt at a Solution



x\sqrt{2x+1} = 0
x must be -1/2 & 0
(-1/2, 0) Critical Point.
(0, 0) Critical Point.

First derivative:
f'g + g'f
(1)(\sqrt{2x+1})+(1/2\sqrt{2x+1})(x)(2)
=(\sqrt{2x+1}) + (x / \sqrt{2x+1})
LCD:
((\sqrt{2x+1}) (\sqrt{2x+1}) + 1)) / (\sqrt{2x+1})

= 2x+1 / \sqrt{2x+1}
So y= 0 when x= -1/2 and 0
(-1/2, 0) Critical Point.

The second derivative I wouldn't post it her since it gets pretty messy. However, I found out that y=o for the second derivative must have x = -1/2

Now the problem is that my answer for all critical points are wrong. What am I doing wrong? PPLEASE HELP!
 
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1calculus1 said:

Homework Statement



Find all critical numbers of f(x)= x\sqrt{2x+1}

Homework Equations



Derivative, Product Rule

The Attempt at a Solution



x\sqrt{2x+1} = 0
x must be -1/2 & 0
(-1/2, 0) Critical Point.
(0, 0) Critical Point.

First derivative:
f'g + g'f
(1)(\sqrt{2x+1})+(1/2\sqrt{2x+1})(x)(2)
=(\sqrt{2x+1}) + (x / \sqrt{2x+1})
LCD:
((\sqrt{2x+1}) (\sqrt{2x+1}) + 1)) / (\sqrt{2x+1})
y=x \sqrt{2x+1}

\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)

\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}

\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0
 
rock.freak667 said:
y=x \sqrt{2x+1}

\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)

\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}

\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0

OH! Thanks for that!
But, the thing is.. the answer for the critical number is -1/3 and from looking at the first derivative, one of the critical point is -1/2.

So what am I doing wrong?
 
While the derivative is undefined at x = -1/2, is the original function differentiable there, and what does that tell you about whether -1/2 is a critical number? Otherwise, if x > -1/2, then you can solve for dy/dx = 0, and you should find x = -1/3. Try it again using the derivative rock.freak posted.
 
Tedjn said:
While the derivative is undefined at x = -1/2, is the original function differentiable there, and what does that tell you about whether -1/2 is a critical number? Otherwise, if x > -1/2, then you can solve for dy/dx = 0, and you should find x = -1/3. Try it again using the derivative rock.freak posted.

OH! Thanks for that. Problem solved.
 
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