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

In summary, Daniel found the coordinates of a stationary point on the curve y=\frac{16x^3 + 4x^2 + 1}{2x^2} and determined the nature of this points.
  • #1
footprints
100
0
Find the coordinates of a stationary point on the curve [tex]y= \frac{16x^3 + 4x^2 + 1}{2x^2}[/tex] And determine the nature of this points.

How I find the coordinates? I know the second part.
 
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  • #2
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
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
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
How do you get the derivitive? When I let me derivitive equal to 0, I keep getting weird answers :grumpy:
 
  • #6
footprints said:
How do you get the derivitive? When I let me derivitive equal to 0, I keep getting weird answers :grumpy:

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
dextercioby said:
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 derivitive [tex]8 - 4x^{-3}[/tex]
Is that correct?
 
  • #8
Yes,it's correct.Now solve the equation into reals...

Daniel.
 
  • #9
maverick280857 said:
Assuming that you mean this by a stationary point

What else would he mean? :confused:
 
  • #10
When [tex]8 - 4x^{-3} = 0[/tex], I get 0.7937... (thats why I said it was weird). I didn't anything wrong did I?
 
  • #11
If your answer is
[tex] x=\frac{1}{\sqrt[3]{2}} [/tex]
,then it is correct.

Daniel.
 
  • #12
Yeah, I got that. However my books answer is [tex](\frac{1}{2}, 8)[/tex]
 
  • #13
You have NOT calculated the derivative correctly (despite what others have told you):
[tex]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}}[/tex]
Hence, your root is [tex]X=\frac{1}{2}[/tex]
 
  • #14
Your book is right, BTW.
 
  • #15
U're right,Arildno,my mistake... :redface:

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

whose derivative is immediate
[tex] y'(x)=8-x^{-3} [/tex]

Daniel.
 
  • #16
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
I don't understand. When differentiating [tex] y(x)=8x+4+\frac{1}{2x^{2}} [/tex], won't I get [tex]8 - 4x^{-3}[/tex]?

[tex]\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(2x^{-2-1})[/tex]
[tex]\frac{dy}{dx}= 8 - 4x^{-3}[/tex]
Whats wrong with that?
Maybe you could show me in detail how it is done?
 
  • #18
footprints said:
I don't understand. When differentiating [tex] y(x)=8x+4+\frac{1}{2x^{2}} [/tex], won't I get [tex]8 - 4x^{-3}[/tex]?

[tex]\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(2x^{-2-1})[/tex]
[tex]\frac{dy}{dx}= 8 - 4x^{-3}[/tex]
Whats wrong with that?
This is wrong; it should be:

[tex]\frac{dy}{dx}= 1*8x^{1-1}+ (-2)(\frac{1}{2}x^{-2-1})[/tex]
 
  • #19
Oh right! Thanks for the help guys!
 

1. What is the purpose of finding stationary points on a function?

The purpose of finding stationary points on a function is to identify the points where the slope or gradient of the function is equal to zero. These points can indicate the maximum or minimum values of the function and are important in understanding the behavior of the function.

2. How do you find the stationary points on a function?

To find the stationary points on a function, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the stationary points. These x-values can then be substituted back into the original function to find the corresponding y-values.

3. What is the significance of maxima and minima in a function?

Maxima and minima are important points in a function as they represent the highest and lowest values that the function can reach. They can provide information about the behavior of the function, such as the optimal value or the turning points.

4. Can a function have more than one stationary point?

Yes, a function can have multiple stationary points. This means that there can be more than one point where the slope or gradient of the function is equal to zero. These points can be classified as either maxima or minima, depending on the behavior of the function.

5. How do you determine if a stationary point is a maximum or a minimum?

To determine if a stationary point is a maximum or a minimum, you can use the second derivative test. If the second derivative is positive, the stationary point is a minimum. If the second derivative is negative, the stationary point is a maximum. If the second derivative is zero, the test is inconclusive and further analysis may be needed.

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