Math Goodies is a free math help portal for students, teachers, and parents.
|
Interactive Math Goodies Software

testing left nav
 Math Forums @ Math Goodies Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums  Homework Help Forums  Algebra  Finding the range

Note: You must be registered in order to post a reply.

Screensize:
Format Mode:
Format:

Message:
* HTML is OFF
* Forum Code is ON

 Math Symbols

Check here to subscribe to this topic.

 T O P I C    R E V I E W moon1130 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 UltraglideAn 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].

 Math Forums @ Math Goodies © 2000-2004 Snitz Communications