Solve f(t) = 3: Finding Domain of f(t) = sqrt(t2 - 16)

[Jacobpm64]

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

 

[d_leet]

Note that $$\sqrt{x^2}\ =\ |x|$$

 

[benorin]

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

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

 

[HallsofIvy]

As you noticed, $t^2 \ge 16$ does not give $t \ge 4$. A standard way of solving inequalities is to solve the corresponding equality first. $t^2= 16$ 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 $x \le -4$ and all $x \ge 4$.
That method works for all continuous functions. Another way to do this inequality is to factor: $t^2- 16= (t-4)(t+4)\ge 0$. 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.