Question

what is the second derivative test, and how can you identify all maximum and minimum values

Answer 1

You take the first derivative in order to identify the critical points and you plug those points into the second derivative in order to determine if it is a max or min.

Answer 2

okay let me try it.

Answer 3

okay using the critical numbers from the first derivative, i get the local max/min..
how would i determine the abs max/min?

Answer 4

In order to find out the max or min you would need to plug in those critical values into the second derivative and if it is negative, then it is a max and if it is positive, then it is a min. Are you given this within a specific interval?

Answer 5

okay, here is the entire question:
use the second derivative test to identify all max and min values of f(x)=x^3+1.5x^2-7x+5 on the interval -4 =< x =< 3

Answer 6

and the answer is:
Local Max (-2.107, 17.054)
Local Min (1.107, 0.446)
Abs Max (3,24.5)
Abs MIn (-4,-7)
i got the local max/min, i dont know how to get the abs max/min

Answer 7

Because the the local max and min do not have larger values than where the interval is located the absolute max and min are located at the intervals which are -4 and 3

Answer 8

but how would you determine that?

Answer 9

Plug the numbers at the interval into the original equation and you will notice that the y values will be larger for the maximums and smaller for the minimums.

Answer 10

oh okay

Answer 11

thanks :)