Relative Max/Min: Solving for Minimum in f(x)

[barbet_psg]

I'm taking AP Calc AB this year and this is a question I've come across while doing work on my own and I won't be seeing my teacher for a few days, so I' figured I'd ask some of you guys. Let's say you have an equation f(x)=(x^3)+(5x^2) and you need to find the relative minimum of that function. Take the derivative and set it to zero to find where the slope levels off, but how do you know which of the two x values gives the minimum? Is it a common sense type thing, because I know that with such a simple function you can deduce that, because it's a positive cubic function, the maximum will come first, and so naturally the minimum will be the larger x value, but is there a more mathematical way to do this? Thanks in advance.

Just kidding, I found out how to do it. Pretty easy.

 

[Vorde]

So we don't need to talk about second derivatives?

 

[HallsofIvy]

You don't need to. The "first derivative test" says that if, at a critical point, the first derivative changes from positive to negative (so the function values are going up, then down), then the critical point is a maximum, if negative to positive, a minimum. Of course, if the first derivative is going from positive to negative, it is an decreasing function and so its derivative, the second derivative of the function, is negative.

 

[middleCmusic]

In the case of a positive cubic function like the above, the relative minimum would always occur at the left endpoint of the interval on which it is being evaluated because it's always increasing. Likewise for the relative maximum. Right?

In general though, you use the second derivative test. (That's the short answer.) But even then, you should check endpoint values and compare.