Domain and Range

  • #1
f(x) = -√(t(1-t))

Domain: 0≤t≥1 (from Null Factor law)
Range: 0≤t

However, the range was wrong. The answers said that it was -0.5≤t≥0. I have no idea where the -0.5 came from. I substituted the domain values in but i didn't work. It just came out with zero.

Can anyone suggest any ideas without the use of minimums?
 

Answers and Replies

  • #2
828
2
f(x) = -√(t(1-t))

Domain: 0≤t≥1 (from Null Factor law)
Range: 0≤t

However, the range was wrong. The answers said that it was -0.5≤t≥0. I have no idea where the -0.5 came from. I substituted the domain values in but i didn't work. It just came out with zero.

Can anyone suggest any ideas without the use of minimums?

Well, there is a minus sign on the front of that radical.

Now, what is the biggest value that the function t(1-t) attains?
 
  • #3
Well, there is a minus sign on the front of that radical.

Now, what is the biggest value that the function t(1-t) attains?

I would say zero.
 
  • #4
828
2
I would say zero.

It isn't zero, but let's assume for the moment that it is zero. Since t(1-t) is under the radical, this would imply that the range of the function is simply 0 since nothing under the radical can be negative. This, in turn, implies that the function t(1-t) must be zero on the interval [0,1]. Clearly, this isn't the case, so 0 is not the highest value the function attains.

What about the midpoint of the interval [0,1]?
 
  • #5
35,000
6,753
f(x) = -√(t(1-t))

Domain: 0≤t≥1 (from Null Factor law)
Quibble about what you wrote for the domain: it should be 0 ≤ t 1 .
 

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