Curve Sketching

  • Thread starter bradycat
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  • #1
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Curve Sketching....

Curve Sketching stuck
Hi,
Stuck on curve sketching on the following.

You have part A which is finding the x-intercepts when y=0.
y = x^5 -5x
0=x^5-5x
x(x^4-5)=0
X=0 and then x^4=5??? stuck here

Then Part b is min/max when y'=0
y'=5x^4-5

5(x^4-1)=0
5=0 CANT USE then X= ROOT of 1 comes to x= - or + 1????
Then to solve for Y it s x= 1,-4 and -1,4 ???

Part C is pts of inflection y''=0
So it's y"=20x^3
Don't know what to do here

Can some one direct me in the right direction, thanks
Joanne
 

Answers and Replies

  • #2
35,019
6,770


Curve Sketching stuck
Hi,
Stuck on curve sketching on the following.

You have part A which is finding the x-intercepts when y=0.
y = x^5 -5x
0=x^5-5x
x(x^4-5)=0
X=0 and then x^4=5??? stuck here
This can be factored some more.
x(x2 - 51/2))(x2 + 51/2) = 0
==> x(x - 51/4)(x + 51/4)(x2 + 51/2) = 0
Then Part b is min/max when y'=0
y'=5x^4-5

5(x^4-1)=0
5=0 CANT USE
This is a bit silly. Of course 5 is not equal to 0. You can divide both sides of the equation by 5, right?
then X= ROOT of 1 comes to x= - or + 1????
Yes, plus two imaginary solutions that you're probably not interested in.
Then to solve for Y it s x= 1,-4 and -1,4 ???
I know what you're trying to say, but you're not doing it very well. If x = 1, y = -4. If x = -1, y = 4. IOW there are critical points at (1, -4) and (-1, 4).

But is either of these a local or global maximum or local or global minimum? There is more you need to do to determine these attributes.
Part C is pts of inflection y''=0
So it's y"=20x^3
Don't know what to do here
What does your book have to say about finding inflection points?
Can some one direct me in the right direction, thanks
Joanne
 
  • #3
13
0


I got it all, I was confusing it with something else, why I was having the problems in the first place.
 

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