Problem+16+Zach+F.+and+Sabrina

 Sabrinbrin: (a)find the intervals on which f is increasing or decreasing (b)find the local maximum and minimum values of f (c)find the intervals of concavity and the inflection points

f(x) = x^2 * ln(x)


 * THE DOMAIN IS x>0

find the derivative (product rule): f'(x) = 2x * ln(x) + x^2/x = 2x * ln(x) + x

set derivative equal to 0 and solve for x: 0 = 2x * ln(x) + x -x = 2x * ln(x) (-1/2) = ln(x) e^(-.5) = e^ln(x) .6065 = x

number line: <-0--.6065--->

use substitution to find where it is increasing/ decreasing (the derivative is pos/neg): 2(.5) * ln(.5) + .5 = -.19315 (NEGATIVE) 2(1) * ln(1) + .5 = .5 (POSITIVE)

(a) (0, .6065) f(x) is decreasing; (.6065, infinity) f(x) is increasing (b) local min at x=.6065

Ms. Sweeney comment: All of this is good, but I think if you were to do this on an AP Exam, it would be good to set up a table: And actually on my test, you may do a number line AS LONG AS you explain your work clearly (again, the important stuff is for you to state that where the derivative is positive, the function is increasing and vice versa)
 * interval || f ' (x) || f ( x) ||
 * (0, 0.6065) || negative || decreasing ||
 * (0.6065, ∞) || positive || increasing ||

find the second derivative: f''(x) = 2*ln(x) + 2x/x + 1 =2ln(x) + 3

set equal to 0: 0 = 2ln(x) + 3 3 = 2ln(x) 1.5 = ln(x) e^1.5 = e^ln(x) x= 4.482

I actually get x = e^-1.5, so you will need to test a value of x that is between 0 and e^-1.5 (pretty small!)

 number line <-4.482> ALWAYS POSITIVE!

(c) always concave up...no point of inflection... (??)

 Zach F. All of my answers agree with yours, however, my answer for number C is different than yours. The mistake that i believe you made is the following find the second derivative: f''(x) = 2*ln(x) + 2x/x + 1 =2ln(x) + 3

set equal to 0: 0 = 2ln(x) + 3 3 = 2ln(x)> here the three should be negative...this changes the answer to the problem drastically 1.5 = ln(x) e^1.5 = e^ln(x) x= 4.482  here is my solution for letter c of the problem c) 2ln(x) +3 = 0 2ln(x) = -3 ln(x) = -1.5 e^ln(x) = e^-1.5 x = e^-1.5 x = .223 at this point i set up a number line that looked like the following 0-.223> then i took a value less than 0.223, lets say 0.1 and plugged it into the 2nd derivative equation 2ln(0.1) + 3 = -4.605 + 3 = negative number then i plugged in a number greater than .223, lets say 1 into the 2nd derivative equation 2ln(1) + 3 = 0 + 3 = positive number Thus i determined that from (0, .223) the original graph is concave down, and from (.223, infinity) the original graph is concave up the x value of the inflection point occurs at e^-1.5 or .223, in order to find the y value i plugged the x value into the original equation (.223)^2(ln(.223) = -.0746 Thus the coordinates of the Inflection point are (.223, -.0746)