Results 1 to 4 of 4

Thread: Nonlinear inequality

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    40

    Nonlinear inequality

    Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set.


    I am stuck on where the line should start, I know it is a positive infinity, it appears as if it starts at -1, but its not correct.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Start by reasoning on the Domain and Range.

    Domain doesn't help much.

    Range?

    $\displaystyle x^{4} \ge 0$

    $\displaystyle x^{9}$ has the same sign as x.

    This restricts your solutions to $\displaystyle x > 0$.

    After that:

    $\displaystyle x^{9} - x^{4} > 0$

    $\displaystyle x^{4} \cdot (x^{5} - 1) > 0$

    Only $\displaystyle x \ne 0$ for $\displaystyle x^{4}$. How about the other piece?

    $\displaystyle x^{5} - 1 = (x-1)(x^{4}+x^{3}+x^{2}+x+1)$

    That big piece restricts nothing, being always greater than zero (0).

    The only piece of any remaining significance is (x - 1).

    This is a great problem to see if you were paying attention in class. Were you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    Edit: sorry, didnt see tkhunny's post. Feel free to ignore this
    $\displaystyle x^9>x^4$
    if $\displaystyle x \not = 0$
    $\displaystyle \frac{x^9}{x^4} >1$
    $\displaystyle x^5>1$
    $\displaystyle (x^5)^{1/5} >1^{1/5}$
    $\displaystyle x>1$

    if x = 0 then $\displaystyle 0^9\not > 0^4$

    I know it is a positive infinity,
    Right
    Last edited by badgerigar; Feb 1st 2009 at 07:14 PM. Reason: TKhunny beat me to it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Well it's like having $\displaystyle x^3>1$ then the cubic root "does the work," since this is just $\displaystyle x>1.$ But this is actually the answer, but it does require a little bit of justification.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] nonlinear ODE
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Oct 20th 2011, 12:12 PM
  2. Nonlinear ODE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Jan 2nd 2011, 07:51 AM
  3. Replies: 3
    Last Post: Dec 12th 2010, 01:16 PM
  4. Nonlinear ODE
    Posted in the Differential Equations Forum
    Replies: 13
    Last Post: Apr 25th 2009, 10:41 AM
  5. nonlinear ODE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Feb 28th 2009, 06:19 AM

Search Tags


/mathhelpforum @mathhelpforum