Solve Curve Sketching Problems with Joanne

[bradycat]

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

 

[Mark44]

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 - 5^1/2))(x² + 5^1/2) = 0
==> x(x - 5^1/4)(x + 5^1/4)(x² + 5^1/2) = 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

 

[bradycat]

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