MHB Concavity and Inflection Points II

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The discussion revolves around solving a calculus problem involving derivatives and asymptotes. The user, ArdentMed, is struggling with the first derivative f'(x) and has calculated critical points and inflection points. A response indicates that ArdentMed's first derivative is incorrect and provides the correct form, suggesting that the critical number is a whole number. Additionally, it confirms that there are no inflection points due to the second derivative being undefined at certain points. The conversation emphasizes the importance of accurately calculating derivatives to determine critical points and asymptotic behavior.
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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.
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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.
 
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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.
 

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