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Critical Number from Analying a Graph.

  • Thread starter 1calculus1
  • Start date
  • #1
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Homework Statement



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

Homework Equations



Derivative, Product Rule

The Attempt at a Solution



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

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

= 2x+1 / [tex]\sqrt{2x+1}[/tex]
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!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31

Homework Statement



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

Homework Equations



Derivative, Product Rule

The Attempt at a Solution



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

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

[tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)[/tex]

[tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}[/tex]

[tex]\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0[/tex]
 
  • #3
39
0
[itex]y=x \sqrt{2x+1}[/itex]

[tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)[/tex]

[tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}[/tex]

[tex]\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0[/tex]
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?
 
  • #4
737
0
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.
 
  • #5
39
0
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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