Mathmate, I really appreciate all your help regarding my composite function question below but I'm still in need of your help. I don't understand how the range of x^2-x+1 is R. Isn't this a parabola with a minimum for y? Also, I understand minimum and maximum values of sine but I don't know how to apply in my question. Again, many thanks for helping me out with this 'cause it's really bugging me.

You're right. The range of an odd-degreed polynomial is ℝ but even degreed polynomials are limited on one side. It's a slip on my part.

In this case, you need to find the minimum (at the vertex) of f(x)=x²-x+1 (1/2,3/4). So the range is [3/4,∞).

To find the minimum and maximum values (range) of g(x)=sin²(x)-sin(x)+1,
you would use differentiation to find the derivative, g'(x)=2sin(x)cos(x)-cos(x)
and equate g'(x) to zero to get
2sin(x)cos(x)-cos(x)=0
cos(x)(2sin(x)-1)=0
thus
cos(x)=0 or sin(x)=1/2
Solution set in [0,2π] is
{π/2, 3π/2, π/6, and 5π/6}
These are the possible locations for absolute minimum/maximum.
Evaluate g(x) at these points and determine the values of the absolute minimum and maximum. These are the limits of the range, since g(x) is a continuous function.

If you have not yet done differential calculus, you can draw the graph of g(x) and select points near which you can calculate a refined value of the absolute maximum and minimum.

Here's graph of the function g(x) between 0 and 2π.

http://img535.imageshack.us/img535/7599/1285459685.png

Thanks very much for taking the time to help! Since I haven't done calculus, I'll try graphing.

You're welcome!

Post if you have other questions, or need more details.

I apologize for the confusion earlier. Let's go step by step to understand the range of the function f(x) = x^2 - x + 1.

To find the range, we need to determine the set of all possible values that the function can take. In this case, we are dealing with a quadratic function, which represents a parabola.

First, let's find the vertex of the parabola. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)). In our case, a = 1, b = -1, and c = 1.

So, the x-coordinate of the vertex, -b/2a, is equal to -(-1)/(2*1) = 1/2. To find the y-coordinate, we substitute this x-value back into the function: f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4.

Therefore, the vertex is (1/2, 3/4), and it represents the minimum point on the parabola.

Now, let's consider the range of the function. Since the parabola opens upward (because the coefficient of x^2 is positive), the parabola has a minimum value, but no maximum value. This means that there is no upper limit to the values that the function can take.

To prove this, we can observe that as x approaches positive or negative infinity, the function approaches positive infinity. Therefore, the range of f(x) = x^2 - x + 1 is (-∞, +∞), which means that it can take any real value.

As for your question about applying the minimum and maximum values of sine, it seems unrelated to the topic of the composite function. If you have any specific questions or examples related to the maximum and minimum values of sine, please feel free to provide more details, and I'll be happy to help you further.