Intervals of Increase and Inflection Points Question

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SUMMARY

The discussion focuses on the differentiation of a function using the quotient rule, specifically addressing the first and second derivatives, as well as the identification of inflection points and asymptotes. The user, ArdentMed, initially computed the first derivative as f'(x) = [2x(x-2)-2(x^2)]/(x-2)^3 but was advised to simplify it to f'(x) = -4x/(x-2)^3 for easier computation of the second derivative. The second derivative was later computed as f''(x) = (8x-12)/(x-2)^2, leading to the identification of inflection points at x = -3/2 and x = 0. The discussion confirms the existence of a vertical asymptote at x = 2 and a horizontal asymptote at y = 1.

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  • Knowledge of first and second derivatives
  • Familiarity with identifying asymptotes in rational functions
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  • Study the process of finding inflection points in detail
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Students studying calculus, particularly those focusing on differentiation, asymptotic behavior, and inflection points. This discussion is beneficial for anyone seeking to improve their understanding of the quotient rule and its applications in analyzing functions.

ardentmed
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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:

This thread is only for question one. Please ignore number two.
08b1167bae0c33982682_22.jpg


So I used the quotient rule to differentiate, giving me:

f'(x) = [2x(x-2)-2(x^2)]/(x-2)^3

Moreover, I proceeded to find f'(x)=0 and f'(x) = DNE, which gave me f(0)=0 and x =/ 2 respectively.

Then I determined concavity by taking f''(x), which gave me 0, albeit I'm not too sure about this one. I ended up getting:
f''(x)= [-4(x^3 - 6x^2 + 11x - 8)]/[(x-2)^6], so I may have made a mistake while applying the quotient rule. I also got f''(x) DNE at x=/2.

For asymptotes, I took lim x-> 2 for the vertical asymptote and got undefined. Therefore, a vertical asymptote exists for x=2, and lim x-> infinity gave me 1, so the horizontal asymptote must be at x=1, correct?

Am I on the right track?


Thanks in advance for all the help guys.

Cheers,
ArdentMed.
 
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Your first derivative is correct, however, you should try to simplify your answer as far as possible. It will make taking your second derivative much easier.
$$f'(x) = \frac{-4x}{(x-2)^3}$$

Retake your second derivative with the simplified first derivative. Your asymptotes are correct.
 
Rido12 said:
Your first derivative is correct, however, you should try to simplify your answer as far as possible. It will make taking your second derivative much easier.
$$f'(x) = \frac{-4x}{(x-2)^3}$$

Retake your second derivative with the simplified first derivative. Your asymptotes are correct.

Thanks for the advice. I took the second derivative and computed:

(8x-12)/(x-2)^2, which seems correct.

Moreover, I found f''(x)=0 and f''(x)= DNE and computed x= -3/2 and x =/ 0 respectively. Are these the inflection points?
 
Actually, your second derivative is also incorrect. Can you find the mistake? Taking the quotient rule, your square your denominator. I'm not sure how you ended up with a power of two. The quotient rule is $$\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$$
where f(x) is $$-4x$$ and g(x) is $$(x-2)^3$$
 

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