# Curve Sketching

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
Mentor

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