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October 23, 2014

October 23, 2014

Posted by **Sarah** on Saturday, September 25, 2010 at 8:08pm.

- Math -
**MathMate**, Saturday, September 25, 2010 at 8:37pmYou'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

- Math -
**Sarah**, Saturday, September 25, 2010 at 11:30pmThanks very much for taking the time to help! Since I haven't done calculus, I'll try graphing.

- Math -
**MathMate**, Sunday, September 26, 2010 at 8:52amYou're welcome!

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

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