Results 1 to 4 of 4

Math Help - stuck on this one problem

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    9

    stuck on this one problem

    Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

    find a and b such that f is differentiable everywhere

    f(x)= ax^3, x=less than or equal to 2
    x^2+b, x=greater than 2

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    You only have problems to differentiate at x=2 so you must find a,b such as f is continuous and differentiable there, i.e. to find values such as \lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x)=f(2) and \lim_{h\to 0}\frac{f(2+h)-f(2)}{h} exists.

    \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}x^2+b=4+b and f(2)=8a so our first condition is 4+b=8a.

    On the other hand, \lim_{h\to 0^{+}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2+b-4-b}{h}=\lim_{h\to 0^{+}}\frac{(2+h)^2-4}{h}=4. and \lim_{h\to 0^{-}}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^{-}}\frac{a(2+h)^3-8a}{h}=4a so our second condition is 4a=4.

    Finally a=1 and b=4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    9
    thanks for showing me how to do it. just one question, could there be another answer because in my text book it is different than what you gave me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by soldatik21 View Post
    Can anyone help me out with this one problem please, didn't really learn this type of problem yet.

    find a and b such that f is differentiable everywhere

    f(x)= ax^3, x=less than or equal to 2
    x^2+b, x=greater than 2

    thanks
    Continuity at x = 2: a(2)^3 = (2)^2 + b => 8a = 4 + b .... (A)

    Differentiability at x =2: 3a(2)^2 = 2(2) => 12a = 4 => a = 1/3.

    Substitute this value of a into equation (A) to get b.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. stuck on this problem for cal 1
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 16th 2008, 06:50 AM
  2. Stuck on a problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 28th 2008, 11:36 AM
  3. here is another problem that i stuck on
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 28th 2007, 08:58 PM
  4. I got stuck on this problem: C(n,6)=C(n,9)
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 7th 2007, 07:22 AM
  5. Stuck on an SAT problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 30th 2006, 12:59 AM

Search Tags


/mathhelpforum @mathhelpforum