# Maxima & Minima

1. Jan 29, 2005

### 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.

2. Jan 29, 2005

### \$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

3. Jan 29, 2005

### 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).

4. Jan 29, 2005

### 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.

5. Jan 29, 2005

### footprints

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

6. Jan 29, 2005

### dextercioby

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.

7. Jan 29, 2005

### footprints

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

8. Jan 29, 2005

### dextercioby

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

Daniel.

9. Jan 29, 2005

### Nylex

What else would he mean?

10. Jan 29, 2005

### footprints

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

11. Jan 29, 2005

### dextercioby

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

Daniel.

12. Jan 29, 2005

### footprints

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

13. Jan 29, 2005

### 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 $$X=\frac{1}{2}$$

14. Jan 29, 2005

### arildno

15. Jan 29, 2005

### 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.

16. Jan 29, 2005

### 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.

17. Jan 29, 2005

### 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?

18. Jan 29, 2005

### arildno

This is wrong; it should be:

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

19. Jan 29, 2005

### footprints

Oh right!!!! Thanks for the help guys!