MHB Intervals of Increase and Inflection Points Question

[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:

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

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.

 

[Dethrone]

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.

 

[ardentmed]

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?

 

[Dethrone]

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 $$(x-2)^3$$