1. Oct 22, 2007

BuBbLeS01

1. The problem statement, all variables and given/known data
Find all relative extrema using the second derivative test for H(x) = x * lnx

2. Relevant equations

3. The attempt at a solution
H'(x) = (1 * ln x) + (x * 1/x) = lnx + 1
H''(x) = 1/x + 0

Is H''(x) right? Then I am having trouble finding the relative extrema from the second derivative test?

2. Oct 22, 2007

eyehategod

h''(x)=1/x

3. Oct 22, 2007

BuBbLeS01

Okay so how do I solve for the relative extrema?

4. Oct 22, 2007

eyehategod

just try graphing the function and go from there. i dont remember what the second derevative test is, its been a long time. dont you have to set denominator =0 and solve x?if so x=0.

5. Oct 22, 2007

Dick

Set H'(x)=0 and solve for x to find the critical points. For each critical point x test the sign of H''(x) to see if it's a max or a min.

6. Oct 23, 2007

BuBbLeS01

H"(x) = 0
I am having trouble finding the relative extrema from the second derivative test?

7. Oct 23, 2007

Dick

Set H'(x)=0 not H''(x)=0. Critical points are where H'(x)=0.

8. Oct 23, 2007

BuBbLeS01

How do you solve lnx + 1 = 0 for x????

9. Oct 23, 2007

BuBbLeS01

would it be...
lnx + 1 = 0
lnx = -1
x = e^-1

10. Oct 23, 2007

BuBbLeS01

Then I would plug that into H" = 1/x, H" = 1/(e^-1) = e > 0 so its a relative max???

11. Oct 23, 2007

Dick

That's almost all exactly correct, except the final conclusion. How can you say H''(x)>0 at a critical point means it's a max?? Don't you have like a textbook or something?

12. Oct 24, 2007

BuBbLeS01

H''(e^-1) > 0 means a relative max...thats what our book says to write? What do you mean? Is that wrong?

13. Oct 24, 2007

HallsofIvy

Either you should throw away your textbook or you should read it more carefully- that's exactly backwards!

14. Oct 30, 2007

GBK.Xscape

Yea, for the second derivative tests things are sort of backwards.
If f"(c)<0, then x=c is a relative maximum.
If f"(c)>0, then x=c is a relative minimum.
f f"(c)=0 or undefined, then the second derivative test is inconclusive.