1. ## function

f)
If g:[1,a] -> R where g(x) = f(x) and g is one-one, write down the largest possible
value of a.

This is part f) of the Q.2, which gives the function $-2x^2+5x-3$ for $x \in [1,\infty)$

2. Does anyone else think $a$ should be the maximum of $f(x)$?

3. Originally Posted by DivideBy0
f)
If g:[1,a] -> R where g(x) = f(x) and g is one-one, write down the largest possible
value of a.

This is part f) of the Q.2, which gives the function $-2x^2+5x-3$ for $x \in [1,\infty)$
I'm just a little confused over all of this. Pretend that I haven't done the problem up to this point. What the heck is f(x) anyway? Are you saying $f(x) = -2x^2 + 5x - 3$?

You have $g: [1, a] \to \mathbb{R}: g(x) = f(x)$. You know that g is one-to-one.

So the question is really: Given the restriction of the domain of f to [1, a], what is the largest value of a such that f(x) is one-to-one.

If your $f(x) = -2x^2 + 5x - 3$ then it looks like x = 1 is on the left of the vertex, so a would have to be the x coordinate of the vertex of f(x). (That's the accurate way to say what I think you were trying to say.)

-Dan

4. Originally Posted by DivideBy0
Does anyone else think $a$ should be the maximum of $f(x)$?
Yes that is the correct answer.
An easy to see it is to simlpy graph the function.

5. Originally Posted by topsquark
I'm just a little confused over all of this. Pretend that I haven't done the problem up to this point. What the heck is f(x) anyway? Are you saying $f(x) = -2x^2 + 5x - 3$?
Thanks both of you
I too was a bit confused over its ambiguity. It didn't say that $-2x^2+5x-3$ was f(x), so I guess it had to be assumed.