Finding the range of f(x)=-√(t(1-t)) without calculus

[Procrastinate]

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?

 

[Robert1986]

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?

 

[Procrastinate]

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.

 

[Robert1986]

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]?

 

[Mark44]

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 .