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Homework Help: Finding the domain.

  1. Aug 23, 2006 #1
    If f(t) = sqrt(t2 - 16), find all values of t for which f(t) is a real number. Solve f(t) = 3.

    All right, I know that I have to find values of t that would make the expression under the radical negative. I was trying to set up an equation to just get the domain.

    t2 - 16 > 0

    this would give me the domain, but it gets weird when i start working:

    t2 > 16
    t > + 4

    but, if it's greater than +4 and -4, then it's just overlapping, and I know that isn't the answer. Actually the domain should be less than -4 and greater than 4 (also equal to those values).

    I just don't know how you get the work to come out to that answer hehe... forgot something somewhere i guess.

    thanks in advance for the help
     
  2. jcsd
  3. Aug 24, 2006 #2
    Note that [tex]\sqrt{x^2}\ =\ |x| [/tex]
     
  4. Aug 24, 2006 #3

    benorin

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    Homework Helper

    From [tex]t^2\geq 16[/tex] when you root it, the positive root produces [tex]t\geq 4[/tex] but taking the negative root counts as multiplying by a negative, so flip the inequality to get [tex]t\leq -4[/tex] so the domain is [tex]t\in (-\infty , -4]\cup [4, +\infty )[/tex] or, equivalently [tex]|t|\geq 4[/tex].
     
  5. Aug 24, 2006 #4

    HallsofIvy

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    Science Advisor

    As you noticed, [itex]t^2 \ge 16[/itex] does not give [itex]t \ge 4[/itex]. A standard way of solving inequalities is to solve the corresponding equality first. [itex]t^2= 16[/itex] gives, of course, t= 4 and t= -4. Those are the only places at which t2 can change sign. It t= -5 (less than -4) then t2= (-5)2= 25 which is greater than 16 so t2> 16 for all x< -4. 0 is between -4 and 4 and of course 02= 0 which is less than 16: t2< 16 for all t between -4 and 4. Finally if t= 5 (larger than 4) then t2= 52= 25 which is again larger than 16. [itex]t^2 \ge 16[/sup] for all [itex]x \le -4[/itex] and all [itex]x \ge 4[/itex].
    That method works for all continuous functions. Another way to do this inequality is to factor: [itex]t^2- 16= (t-4)(t+4)\ge 0[/itex]. If t< -4 both factors are less than 0; the product of two negative numbers is positive. If t is between -4 and 4, one factor, t-4, is still negative but the other, t+4 is now positive. The product of a negative number with a positive number is negative. Finally, if t>4, both factors are positive so the product is positive again.
     
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