Find Stationary Point on y= 16x³ + 4x² + 1/2x² | Maxima & Minima

[footprints]

Find the coordinates of a stationary point on the curve $$y= \frac{16x^3 + 4x^2 + 1}{2x^2}$$ And determine the nature of this points.

How I find the coordinates? I know the second part.

 

[$id]

You have to find the dy/dx of that equation and find points where that is 0

Hint: use the quotient rule to differentiate that

 

[arildno]

You say you know about the second part.
We are therefore, I hope, in agreement that the x-values for stationary points are found by the equation y'(x)=0.
Let us denote a particular solution of this equation by X.
But then, the corresponding y-value for a stationary point on the curve (for which the x-value is X) is simply y(X).

 

[maverick280857]

Assuming that you mean this by a stationary point, set the first derivative equal to zero.

EDIT--Oops...when I was replying to this post I didn't see others' responses. Anyway you have much better ones now. Cheers.

 

[footprints]

How do you get the derivative? When I let me derivative equal to 0, I keep getting weird answers

 

[dextercioby]

How do you get the derivative? When I let me derivative equal to 0, I keep getting weird answers

What do you mean by "weird"??Should they be "lovely"?

On the other hand i don't see a really "nice" cubic...
BTW,because "x=0" is not in the domain of the function,u can simplify the quartic on the numerator and end up with a cubic...

Daniel.

 

[footprints]

BTW,because "x=0" is not in the domain of the function,u can simplify the quartic on the numerator and end up with a cubic...

Daniel.

I did that the first time I did it. I got the derivative $$8 - 4x^{-3}$$
Is that correct?

 

[dextercioby]

Yes,it's correct.Now solve the equation into reals...

Daniel.

 

[Nylex]

Assuming that you mean this by a stationary point

What else would he mean?

 

[footprints]

When $$8 - 4x^{-3} = 0$$, I get 0.7937... (thats why I said it was weird). I didn't anything wrong did I?

 

[dextercioby]

If your answer is
$$x=\frac{1}{\sqrt[3]{2}}$$
,then it is correct.

Daniel.

 

[footprints]

Yeah, I got that. However my books answer is $$(\frac{1}{2}, 8)$$

 

[arildno]

You have NOT calculated the derivative correctly (despite what others have told you):
$$y'(x)=\frac{2x^{2}(3*16x^{2}+8x)-4x(16x^{3}+4x^{2}+1)}{4x^{4}}=\frac{2x}{4x^{4}}*((3*16x^{3}+8x^{2})-(2*16x^{3}+8x^{2}+2))=\frac{16x^{3}-2}{2x^{3}}=8-\frac{1}{x^{3}}$$
Hence, your root is $$X=\frac{1}{2}$$

 

[arildno]

Your book is right, BTW.

 

[dextercioby]

U're right,Arildno,my mistake...

The function,IIRC is
$$y(x)=8x+4+\frac{1}{2x^{2}}$$

whose derivative is immediate
$$y'(x)=8-x^{-3}$$

Daniel.

 

[arildno]

footprints:
Take this as a typical example of why I HATE the differentiation rule for fractions!
It is the nastiest one, it is so easy to make a mistake.

Daniel has kindly provided you with a rewriting which gives you the correct answer right away.

 

[footprints]

I don't understand. When differentiating $$y(x)=8x+4+\frac{1}{2x^{2}}$$, won't I get $$8 - 4x^{-3}$$?

$$\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(2x^{-2-1})$$
$$\frac{dy}{dx}= 8 - 4x^{-3}$$
Whats wrong with that?
Maybe you could show me in detail how it is done?

 

[arildno]

I don't understand. When differentiating $$y(x)=8x+4+\frac{1}{2x^{2}}$$, won't I get $$8 - 4x^{-3}$$?

$$\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(2x^{-2-1})$$
$$\frac{dy}{dx}= 8 - 4x^{-3}$$
Whats wrong with that?

This is wrong; it should be:

$$\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(\frac{1}{2}x^{-2-1})$$

 

[footprints]

Oh right! Thanks for the help guys!