Identify the critical points and find the maximum and minimum value on the given interval
F(x)=x^2+4x; I=(-5/2, -1/2)

Question

psymon · Answer

To start off, do we know how to do our derivatives?

anonymous · Answer

Yes. f'(x) = 2x+4

psymon · Answer

Alright, well critical points are found by setting the derivative equal to 0. So solving for x will get you critical points.

anonymous · Answer

So x=-2

psymon · Answer

Right. Now you have to take the critical points and the end points of your interval and plug them into the original equation. From there you'll see which of the 3 points gives you a maximum value and a minimum value and those points will be your answer.

anonymous · Answer

Would I plug it into the original equation or the derivative?

psymon · Answer

The original. Because what you're basically doing is finding the y-coordinate on your graph and see if its the highest point or the lowest point.

anonymous · Answer

And do the points in the interval also count as critical points? Cause I have the answer and that is what is listed along with the -2

psymon · Answer

Only the critical points you find from setting the derivative equal to 0 and the endpoints. No pointsin between count.

anonymous · Answer

So the endpoints would be -5/2 and -1/2?

psymon · Answer

Right. So now you would plug -2, -5/2, and -1/2and see which ones give you the highest answer and the lowest answer.

anonymous · Answer

Okay, thank you so much!

psymon · Answer

yep, np ^_^

anonymous · Answer

I don't know if you're still there but would the derivative for 2 cos theta be -2 sin theta? and if yes how would I set that to zero?

psymon · Answer

Yeah, 2cosx would be -2sinx. The -2 means nothing really for setting the equation to 0. Now its just knowing your unit circle well enough to know where sinx = 0.

anonymous · Answer

pi and 2pi. The -2 was confusing me lol, thanks!

psymon · Answer

Yeah, lol. It doesnt matter because once sin x = 0, 0 * -2 is still 0.