Posted - 03/26/2013 : 23:39:35 I am asked to find the range of the function h(t)=(9-t). I solved the problem by first finding the domain of h, which is -3t3, and then calculating the range from this via a table. I am sure that there is another way to solve this problem algebraically by dealing directly with the range concept (and bypassing the domain), but I have not been able to find it. Can someone help me?

Thank you.

6 L A T E S T R E P L I E S (Newest First)

Ultraglide

Posted - 04/02/2013 : 18:02:40 Oops, you're right it is a semi-circle but everything else is the same.

Subhotosh Khan

Posted - 03/31/2013 : 12:28:29

quote:Originally posted by Ultraglide

An alternate way of looking at this problem is by graphing the function. First note that y=9-t is a parabola opening downward with t-intercepts of 3 and an h-intercept of 9. This function, though, is a square root function, so you are restricted to positive values, hence any part of the function below the t-axis is eliminated. So now you only have h-values between 0 and 9 so if you take the root, you get values between 0 and 3.

The function is h(t) = (9 -t)

Thus it is the upper half of a circle

h = 9 - t → h + t = 3

moon1130

Posted - 03/30/2013 : 23:47:10 I thank each of you for your responses and help. I now understand the process. I cannot put my finger on what exactly confused me earlier, but it is all clear in my mind now.

My answer to this question is [0,3] also. I determined the range by first finding the domain and them calculating the h(t) values for the minimum and maximum range values, respectively.

I was looking for an algebraic method akin to finding the inverse of a one-to-one function.

All of you have a very Good one...

moon1130

Ultraglide

Posted - 03/27/2013 : 11:02:59 An alternate way of looking at this problem is by graphing the function. First note that y=9-t is a parabola opening downward with t-intercepts of 3 and an h-intercept of 9. This function, though, is a square root function, so you are restricted to positive values, hence any part of the function below the t-axis is eliminated. So now you only have h-values between 0 and 9 so if you take the root, you get values between 0 and 3.

the_hill1962

Posted - 03/27/2013 : 10:39:06 The way that you stated is really is a good way to solve it. There is no "algebraic" method. (9-t) does not simplify anyway. For these type of problems, looking for discontinuities (holes or asymptotes, etc) is the way answer it. Sometimes a table might not show the discontinuities. Please let us know what you list for the answer to this problem so we can check to make sure you did it correctly.

royhaas

Posted - 03/27/2013 : 08:22:11 By definition, the range is the set of values corresponding to the domain. In this case, it's [0,3].