For what values y is y(t) increasing/decreasing?

[Duderonimous]

Homework Statement

y'=y^{3}-y^{2}-12y

For what values of y is y(t) increasing and for what values is it decreasing?

Homework Equations

The Attempt at a Solution

I think you take the second derivative and equal it to zero to figure out the inflection pints right and then I am not so sure from there.

y''=3y^{2}-2y-12

0=3y^{2}-2y-12

Quadratic formula

y=\frac{2\pm2\sqrt{37}}{6}

Am I correct in this approach? I don't think I am.

 

[skiller]

Homework Statement

y'=y^{3}-y^{2}-12y

For what values of y is y(t) increasing and for what values is it decreasing?

For y to be either increasing or decreasing, you just need to look at whether y' is positive or negative, so the original equation gets you half-way there already - no need to differentiate again.

Try to factorize the RHS and it should become clear.

 

[rude man]

You also need to take y''. Solving for y' = 0 just gets you the values of y for which the slope is zero. To also learn whether those points are (relative) minima or maxima requires taking y'' and substituting the values of y for which y' = 0.

Yes, factor by all means.

 

[skiller]

You also need to take y''. Solving for y' = 0 just gets you the values of y for which the slope is zero. To also learn whether those points are (relative) minima or maxima requires taking y'' and substituting the values of y for which y' = 0.

Yes, factor by all means.

Rubbish.

 

[rude man]

Rubbish.

I say! I think you're right!

 

[HallsofIvy]

Odd that oay should be ruder than rude man and then rude man agrees with him! But oay is right- the second derivative is irrelevant. You are NOT asked to find "inflection points".

y'= y³- y²- 12y= y(y²- y- 12)= y(y- 4)(y+ 3), then if y< -3, all three factors are negative so y' is negative. If -3< y< 0 then y+3 is positive while y and y= 4 are both negative so y' is positive, etc.

But you would still need to find what values of t give those values of y.

 

[rude man]

I agree with hallsofivy. When you're wrong... admit it! And I do.

 

[skiller]

But you would still need to find what values of t give those values of y.

Finding values of t wasn't in the question, though.

I agree with hallsofivy. When you're wrong... admit it! And I do.

And I apologize for my sharp reply of "Rubbish". I think I'd just had a bottle of Chardonnay.

 

[rude man]

Finding values of t wasn't in the question, though.

And I apologize for my sharp reply of "Rubbish". I think I'd just had a bottle of Chardonnay.

No apology required! It was just that.

 

[Dick]

No apology required! It was just that.

Apt comment gracefully accepted and, shall I say, "wittily" replied to. I worry a lot less about rudeness when replying to somebody with 1600+ posts under their belt. I assume that's made them at least a little bit thick skinned. Nice exchange. I don't think there was any real rudeness intended and better yet, none recieved.