Results 1 to 3 of 3

Math Help - Help please a question

  1. #1
    key
    key is offline
    Newbie
    Joined
    Sep 2007
    Posts
    21

    Help please a question

    Hey. Can someone help me with these questoins

    1. What is the domain of h(x)= square root of x squared - 4x - 5

    How do you find the domain of things I know it's X but how do you find it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Quote Originally Posted by key View Post
    1. What is the domain of h(x)= square root of x squared - 4x - 5
    We require that x^2-4x-5\ge0\implies(x+1)(x-5)\ge0.

    Do you know how to tackle that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by key View Post
    Hey. Can someone help me with these questoins

    1. What is the domain of h(x)= square root of x squared - 4x - 5

    How do you find the domain of things I know it's X but how do you find it
    Well, you can't take the square root of a negative number , so everything inside the square root, must be positive. (if you have difficulty seeing this, ask yourself what real number multiplied by itself will return a negative number? for example what number squared would be negative 1? There is not one, because a negative number squared will become positive, and a positive number squared will remain positive)

    So
    0 \leq x^{2}+2x5

    Now, if h(x) is positive, but was negative (or vice versa), it must have been equal to zero at some point, because zero is where signs change. So lets find where this equation equals zero.
    0 = x^{2}-4x-5

    Factor into (x-1) and (x+1)
    0 = (x-5)(x+1)

    so x-5=0 ~~~\Longrightarrow ~~x=5
    and x+1=0 ~~\Longrightarrow ~~x=-1

    Now, we know where the function equals zero, which is where it changes signs. We just need to check the intervals.

    The intervals are (-\infty,-1), (-1,5), \mbox{ and } (5,\infty)

    So lets pick the point -2 to check the sign of the first interval, 0 to check the sign of the second interval, and 6 to check the sign of the third interval.

    let g(x) = x^{2}-4x-5
    g(-2)=7 ~~\Longrightarrow ~~(-\infty,-1] is in the domain
    g(0)=-5 ~~\Longrightarrow ~~(-1, 5)] is not in the domain
    g(6)=~7 ~~~\Longrightarrow ~~[5, \infty) is in the domain

    (note, we used brackets on -1 and on 5, because they are roots, which means they cause the equation to equal zero, and we can take the square root of zero, so they are in the domain)

    So the domain is: (-\infty, -1]\cup[5, \infty)

    -----
    Now you try for h(x)=\sqrt{-x^{2}-2x+8}
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum