Concavity and Inflection Points II

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question.

I'm only asking about two. Please ignore question one.

f'(x) = 2(x^ 1/3) - 1.

Moreover, I proceeded to find f'(x)=0 and f'(x) = DNE, which gave me f(2^1/3)=4.74 (I highly doubt that this is right).

As for inflection points, since f''(x) = (-2/3)x^(-4/3), no inflection points exist since x is only in the denominator.
For asymptotes, I took lim x-> 0 for the vertical asymptote and got x=0. . Therefore, a vertical asymptote exists at x=0, and lim x-> infinity gave me infinity, so there is no horizontal asymptote.

Am I close?

Thanks in advance for all the help guys.

Cheers,
ArdentMed.

 

[Dethrone]

Your first derivative is wrong, the critical number is actually a whole number :)
$$f(x) = 3x^{\frac{2}{3}} - x$$

Applying the power rule:
$$f'(x) = 3(\frac{2}{3})x^{\frac{-1}{3}} - 1$$

I'll leave the simplifying for you to do.