- #1
ardentmed
- 158
- 0
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.
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.