Results 1 to 4 of 4

Math Help - Find the difference quotient and simplify

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    28

    Find the difference quotient and simplify

    Find the difference quotient and simplify answer: f(5+h)-f(5)/h, h cannot = 0
    when f(p)=6p^2-4p.




    ok this is what i did and i got stuck.

    f(5+h)= 6(5+h)^2-4(5+h)

    =6(25+10h+h^2)-20-4h

    =150+60h+6h^2-20-4h

    = 130+56h+6h^2

    ??????? help me out...its getting so late can't keep my eyes open lol
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by refresh View Post
    Find the difference quotient and simplify answer: f(5+h)-f(5)/h, h cannot = 0
    when f(p)=6p^2-4p.




    ok this is what i did and i got stuck.

    f(5+h)= 6(5+h)^2-4(5+h)

    =6(25+10h+h^2)-20-4h

    =150+60h+6h^2-20-4h

    = 130+56h+6h^2

    ??????? help me out...its getting so late can't keep my eyes open lol
    You're right so far. Now find f(5), substitute both f(5), \ f(5+h) into the formula and simplify.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    The function f is defined as follows :

    f(p) = 6p^2 - 4p

    Basically, you are trying to simplify this :

    \frac{f(5+h)-f(5)}{h} , h \neq 0

    It is equivalent to (by substituting the function into your expression) :

    \frac{[6(5 + h)^2 - 4(5 + h)] - [6(5)^2 - 4(5)]}{h} , h \neq 0

    Expand this :

    \frac{[6(5^2 + 10h + h^2) - 20 - 4h] - (150 - 20)}{h} , h \neq 0

    Further expansion required ...

    \frac{150 + 60h + 6h^2 - 20 - 4h - 130}{h} , h \neq 0

    Simplify :
    > get rid of the constants (150, 130, 20, ...), they cancel out (who would have guessed it ?).
    > simplify the linear coefficients (60h - 4h) !
    > reorganize the equation, looks better ...

    \frac{6h^2 + 56h}{h} , h \neq 0

    Factorize with h :

    \frac{h(6h + 56)}{h} , h \neq 0

    Finally, cancel out the h, to find :

    6h + 56

    And you are done

    EDIT : ah Defunkt you win, I was busy checking my reply :/
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2009
    Posts
    28
    Quote Originally Posted by Bacterius View Post
    The function f is defined as follows :

    f(p) = 6p^2 - 4p

    Basically, you are trying to simplify this :

    \frac{f(5+h)-f(5)}{h} , h \neq 0

    It is equivalent to (by substituting the function into your expression) :

    \frac{[6(5 + h)^2 - 4(5 + h)] - [6(5)^2 - 4(5)]}{h} , h \neq 0

    Expand this :

    \frac{[6(5^2 + 10h + h^2) - 20 - 4h] - (150 - 20)}{h} , h \neq 0

    Further expansion required ...

    \frac{150 + 60h + 6h^2 - 20 - 4h - 130}{h} , h \neq 0

    Simplify :
    > get rid of the constants (150, 130, 20, ...), they cancel out (who would have guessed it ?).
    > simplify the linear coefficients (60h - 4h) !
    > reorganize the equation, looks better ...

    \frac{6h^2 + 56h}{h} , h \neq 0

    Factorize with h :

    \frac{h(6h + 56)}{h} , h \neq 0

    Finally, cancel out the h, to find :

    6h + 56

    And you are done

    EDIT : ah Defunkt you win, I was busy checking my reply :/
    omg i thought i had to substitute the f(5) into 130+56h+6h^2 lol no wonder i been getting the wrong answer. thank you so much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 4th 2012, 11:26 AM
  2. Replies: 4
    Last Post: August 31st 2010, 06:08 PM
  3. Find the difference quotient and simplify answer
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: November 9th 2009, 11:14 PM
  4. Difference Quotient
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 7th 2009, 08:46 PM
  5. simplify the difference quotient
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 4th 2008, 09:27 PM

Search Tags


/mathhelpforum @mathhelpforum