Results 1 to 14 of 14

Math Help - Calculus Assignment - HELP!!!

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    13

    Calculus Assignment - HELP!!!

    If anyone could help me with these questions, that would be really really great.

    1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.

    2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.

    3. Determine lim x-->0 (3rdroot(x-8)+2)/x

    4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.


    Anything you can add will be majorly helpful.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    2. The graph of has a f(x)=(ax+b)/((x-1)(x-4)) horizontal tangent line at (2,-1). Determine the values a and b.
    f(x)=\frac{ax+b}{(x-1)(x-4)}

    Setting f(x) =-1 and subbing in x=2, we get f(x)=-a-\frac{b}{2}

    f'(x)=\frac{-ax^{2}+2bx-4a-5b}{(x-1)^{2}(x-4)^{2}}

    Sub in x=2 into f'(x), we get f'(x)=\frac{b}{4}

    Since the line is horizontal, we have \frac{b}{4}=0

    We have two small equations in 2 unknowns:

    -a-\frac{b}{2}=-1

    \frac{b}{4}=0

    Solve for a and b



    Quote Originally Posted by philipsach1 View Post
    4. The tangent line to the curve defined by g(x)=(x^4-2x^3+3)/(-6x) at x=-1 intersects the curce at two other points P and Q. Determine the coordinates of P and Q, algebraically. Illustrate this situation with a graph.


    Anything you can add will be majorly helpful.

    f(x)=\frac{x^{4}-2x^{3}+3}{-6x}

    f'(x)=\frac{-3x^{4}+4x^{3}+3}{6x^{2}}

    Sub x=-1 into f(x) and we find y=1

    Sub x=-1 into f'(x) to find the slope at x=-1. We get m=-2/3

    Use the given coordinates and the new found slope to find the equation of our line:

    y=mx+b

    1=(\frac{-2}{3})(-1)+b

    b=\frac{1}{3}

    Therefore, the equation of the line is y=\frac{-2}{3}x+\frac{1}{3}

    Set f(x) and the line equation equal and solve for x:

    \frac{x^{4}-2x^{3}+3}{-6x}=\frac{-2}{3}x+\frac{1}{3}

    3x^{4}-6x^{3}-12x^{2}+6x+9=0

    3(x+1)^{2}(x-1)(x-3)
    Last edited by galactus; November 24th 2008 at 05:39 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2006
    Posts
    13

    ???

    Does anyone else got anything for the first three questions???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by philipsach1 View Post
    1. Determine the derivative of f(x)=(2x^2-1)/x from first principles. Use the quotient rule to verify your answer.
    f(x) = \frac{2x^2 - 1}{x}

    f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

    f'(x) = \lim_{h \to 0} \frac{ \frac{2(x+h)^2 - 1}{x + h} - \frac{2x^2 - 1}{x} }{h}

    f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 - 1)x - (2x^2 - 1)(x + h)}{hx(x+h)}

    f'(x) = \lim_{h \to 0} \frac{2x^3 + 4x^2h + 2xh^2 - x - 2x^3 - 2x^2h + x + h}{hx(x+h)}

    f'(x) = \lim_{h \to 0} \frac{2x^2h + 2xh^2 + h}{hx(x+h)}

    f'(x) = \lim_{h \to 0} \frac{2x^2 + 2xh + 1}{x(x+h)}

    f'(x) = \frac{2x^2 + 1}{x^2}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2006
    Posts
    13
    Thank you soooo much!

    If I could just get an answer for number three, I will be the happiest person!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by philipsach1 View Post
    3. Determine lim x-->0 (3rdroot(x-8)+2)/x
    \lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x}

    This is in the form of \frac{0}{0} so use L'Hopital's rule:

    \lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}

    = \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2006
    Posts
    13

    Sorry

    Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by philipsach1 View Post
    Sorry again, but could you explain to me how you used the l'Hopital rule in question 3? i can't figure out the process.
    No problem!

    Note that, for x = 0 the numerator becomes:
    \sqrt[3]{0-8} + 2 = \sqrt[3]{-8} + 2 = -2+ 2 = 0

    So for x = 0 the fraction takes on the form \frac{0}{0}. This is one of the conditions under which we may use L'Hopital's rule.

    L'Hopital's rule says that:
    \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} when \lim_{x \to a}f(x) = 0 and \lim_{x \to a}g(x) = 0.

    So:
    \lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} \to \frac{0}{0}

    We need to take the derivative of both the numerator and denominator.

    Numerator:
    \frac{d}{dx}(\sqrt[3]{x-8} + 2 ) = \frac{d}{dx}((x - 8)^{1/3} + 2)

    = \frac{1}{3}(x - 8)^{-2/3} = \frac{1}{3\sqrt[3]{(x - 8)^2}}

    Denominator:
    \frac{d}{dx}x = 1

    So:
    \lim_{x \to 0} \frac{ \sqrt[3]{x-8} + 2 }{x} = \lim_{x \to 0} \frac{ \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x-8)^2}} }{1}

    = \frac{1}{3} \cdot \frac{1}{\sqrt[3]{(-8)^2}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I'm sorry, correct me if I'm wrong but (0-8)^{\frac{1}{3}}=1+\sqrt{3}i

    I ran this through the calc to be sure, but I get an undefined limit.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Hello,

    have a look here: http://www.mathhelpforum.com/math-he...jmailloux.html

    Are you both in the same class?

    EB
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by galactus View Post
    I'm sorry, correct me if I'm wrong but (0-8)^{\frac{1}{3}}=1+\sqrt{3}i

    I ran this through the calc to be sure, but I get an undefined limit.
    Hello, galactus,

    (0-8)^{\frac{1}{3}}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2

    Quote Originally Posted by galactus View Post

    I ran this through the calc to be sure, but I get an undefined limit.
    The limit should be calculated with x approaching zero.

    EB
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by galactus View Post
    I ran this through the calc to be sure, but I get an undefined limit.
    I'm curious. The screenshot looks like a TI-92, but mine will calculate the limit. What calculator are you using?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I also have a 92. I used x approaching 0. It looks funny in the post.

    My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.
    Last edited by galactus; November 27th 2006 at 05:42 AM.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by galactus View Post
    I also have a 92. I used x approaching 0. It looks funny in the post.

    My apologies. I had the complex format on my calculator set to rectangular instead of real. That made the difference.
    No apologies necessary.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 25th 2010, 10:41 PM
  2. first calculus assignment of the year
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 13th 2009, 01:30 PM
  3. Calculus assignment
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2008, 05:07 PM
  4. Just a Few Calculus Assignment Questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 14th 2007, 12:52 AM
  5. Major Calculus Assignment Issue
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 1st 2006, 06:15 PM

Search Tags


/mathhelpforum @mathhelpforum