Concavity and the second derivative

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Question from Christina, a student:

I'm having trouble solving for a second derivative for the following graphing question.

f(x) = (X^2+2x+4)/2x

using the quotient rule, I found:
f'(x) = (x^2-4)/(2x^2)

however, using the quotient rule again I can't seem to solve it (concavity):
f'''(x)=[(2x)(2x^2)-(x^2-4)(4x)]/[(2x^2)^2]
f''(x)=[(4x^3-(4x^3 -16x)]/4x^4
f''(x)=16x/4x^4
f''(x)=4/x^3

and making the equation equal to zero result in 0=4 which doesn't seem to make sense...

Christina,

Your differentiation is fine and you found f''(x) = 4/x³. Then you attempted to solve f''(x) = 0 and ended with a statement that makes no sense, so there is no x so that f''(x) = 0. But the concavity might change at a place where f''(x) = 0 but also at a place where f''(x) is undefined. f''(x) = 4/x³ is undefined at x = 0 so if the concavity changes it does so at x = 0.

Penny

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