Critical Number from Analying a Graph.

[1calculus1]

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!

 

[rock.freak667]

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)) / ($$\sqrt{2x+1}$$)

$y=x \sqrt{2x+1}$

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

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

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

 

[1calculus1]

$y=x \sqrt{2x+1}$

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

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

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

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?

 

[Tedjn]

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.

 

[1calculus1]

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.