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Question 965413: R(t)=3sqrtt^2-44t+1
How do you find the domain
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the domain is the possible values of x that will result in a valid y.
your equation is:
R(t)=3 * sqrt (t^2-44t+1)
set r(t) equal to y.
equation becomes:
y = 3 * sqrt (t^2 - 44t + 1)
replace t with x because x is easier to work with and to graph.
equation becomes:
y = 3 * sqrt (x^2 - 44x + 1).
i believe that's what you meant.
the parentheses are important to show what epression you are taking the square root of.
this equation will only yield a real answer if x^2 - 44x + 1 is greater than or equal to 0.
otherwise the solution will contain an imaginary component and so will not be considered real.
an imaginary component will occur if you have the square root of a negative number.
so, in your domain, x^2 - 44x + 1 must be greater than or equal to 0.
you need to solve the quadratic equation to find the 0 points and then you need to find the intevals where the equation is greater than 0 and where the equation is less than 0.
factor x^2 - 44x + 1 using the quadratic formula to get:
x = 43.97726098 or x = 0.02273902416
if x is less than .0227....., then the equation will be positive.
if x is greater than 43.9772...., then the equation will be positive.
otherwise, the equation will be negative.
the graph of that equation is shown below:
your domain is the values of x that will lead to a good solution for y.
a good solution for y is a real solution.
this occurs only when x is less than or equal to 0.02273902416 and when x is greater than or equal to 43.97726098.
here's the graph of y = 3 * sqrt(x^2 - 44x + 1)
as you can see, you only get a real solution when x < .0227... and when x > 43.9772...
between those values, y is not real because you have a square root of a negative number.
you can't graph an imaginary number in the real plane, so nothing is shown.
that's why the grpahis empty between the zero points of x^2 - 44x + 1.
a word about functional notation.
your problem startede as f(t) = 3 * sqrt(t^2 - 44T + 1).
i changed t to x.
your problem becamse f(x) = 3 * sqrt(x^2 - 44x + 1).
same sunction, only a different argument.
i also made y = f(x).
this was done for grpahing purposes because x-axis is horizontal and y-axis is vertical.
this satisfies graphing software.
remember:
your domain is the values of x that will lead to a valid solution of y.
with f(t), then your domain is the values of t that will lead to a valid solution of f(t).
here's another example of a domain problem where not all value of x will lead to a good solution for y.
y = 5/x.
your domain is all values of x with the exception of x = 0 because, when x = 0, the value of y is undefined.
undefined is not a valid value of y.
any other value of x will lead to a valid value of y except when x = 0.
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