Originally Posted by

**larasanut** I have a few problems that i don't understand the answer..

f(x)=2x^2 -3x+4

f'(x)= 4x-3

f"(x)=4

answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

Mr F says: Because the domain of f is all real numbers and f'' > 0 for all real numbers.

f(x)=x^4/7

f'(x)= 4/7 x^-3/7

f"(x)=-12/49 x^-10/7

answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

Mr F says: Because the domain of f is all real numbers and f'' < 0 for all real numbers except x = 0 (where it's undefined).

f(x)= root 4-x = (4-x)^1/2

f'(x)=1/2 (4-x)^-1/2

f"(x)= -1/4 (4-x)^-3/2

amswer s cd (-infinity,4)......if i plug in say f(5) i'd get a positive Mr F says: That's strange. Especially since x = 5 is not in the domain of f since f(5) is not defined. Concavity can obviously only exist where the fuction exists, that is, over the domain of the function.

..so why is it not cd(-infinity,4) AND cu (4, infinity)?

thanks a bunch...i really appreciate u guys helping me, i only have bits and pieces of this stuff..but i'm getting the bigger picture a little at a time