Results 1 to 8 of 8

Thread: evaluate and simplify

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    evaluate and simplify

    hey guys, need help with a practise question... thank you!!

    if
    $\displaystyle
    f(x) = \frac{2}{3x+4}
    $

    evaluate and simplify

    $\displaystyle
    \frac{(2+h)-f(2)}{h}
    $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by jvignacio View Post
    hey guys, need help with a practise question... thank you!!

    if
    $\displaystyle
    f(x) = \frac{2}{3x+4}
    $

    evaluate and simplify

    $\displaystyle
    \frac{(2+h)-f(2)}{h}
    $
    have you wriiten the question correctly?

    first, $\displaystyle f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}$

    so, $\displaystyle
    \frac{(2+h)-f(2)}{h} = \frac{(2+h) + \frac{1}{5}}{h} = \frac{10 + 5h + 1}{5h} = \frac{11+5h}{5h}
    $


    but this looks familiar.. i was thinking that the problem was

    $\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$

    am i right?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by kalagota View Post
    have you wriiten the question correctly?

    first, $\displaystyle f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}$

    so, $\displaystyle
    \frac{(2+h)-f(2)}{h} = \frac{(2+h) + \frac{1}{5}}{h} = \frac{10 + 5h + 1}{5h} = \frac{11+5h}{5h}
    $


    but this looks familiar.. i was thinking that the problem was

    $\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$

    am i right?
    yes sorry thats it! your right. thats the problem. how would i solve it like that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    it should be familiar to you that

    $\displaystyle f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

    if not yet, all you have to do is to substitute $\displaystyle 2+h$ on $\displaystyle x $ of $\displaystyle f(x)$ to get $\displaystyle f(2+h)$

    then solve for that limit..

    $\displaystyle f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}$

    so, $\displaystyle \lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}$...

    i hope, you can continue from here..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by kalagota View Post
    it should be familiar to you that

    $\displaystyle f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

    if not yet, all you have to do is to substitute $\displaystyle 2+h$ on $\displaystyle x $ of $\displaystyle f(x)$ to get $\displaystyle f(2+h)$

    then solve for that limit..

    $\displaystyle f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}$

    so, $\displaystyle \lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}$...

    i hope, you can continue from here..
    okay i got upto..

    $\displaystyle
    \lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
    $

    am i right? not sure what to do after
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517
    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    $\displaystyle
    \lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
    $

    am i right? not sure what to do after

    anyone?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    $\displaystyle
    \lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
    $

    am i right? not sure what to do after
    that should be negative..


    $\displaystyle
    \lim_{h\rightarrow 0} \frac{\frac{-3h}{50+15h}}{h} = \lim_{h\rightarrow 0} \frac{-3h}{50+15h} \cdot \frac{1}{h} = \lim_{h\rightarrow 0} \frac{-3}{50 + 15h} ...
    $
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    $\displaystyle
    \lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
    $

    am i right? not sure what to do after
    $\displaystyle
    -\ \frac{\frac{3h}{50+15h}}{h}=-\ \frac{3}{50+15h}
    $

    Then the limit of this as $\displaystyle h$ goes to $\displaystyle 0$ is obtained just by putting $\displaystyle h=0$, so is: $\displaystyle -3/50$.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Aug 31st 2010, 05:08 PM
  2. Evaluate the sum
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 26th 2009, 04:45 PM
  3. Evaluate then simplify
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 26th 2009, 08:06 AM
  4. simplify and evaluate - not limit.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Aug 19th 2008, 11:11 PM
  5. evaluate or simplify without using a calulator
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Mar 31st 2008, 01:06 PM

Search Tags


/mathhelpforum @mathhelpforum