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August 19th, 2012, 12:00 AM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  how to do it algebraically?
Let's consider the following problem. The range of the function y = square root of f(x) is y = > 4. If f(x) = 0.5*(x + 1)^2 + k. What is k? The vertex of this parabola is (1,k). Since the minimum value of the range of the square root function is y=4, we can square it to find the y coordinate of the parabola's vertex. So, the answer is k=16. I am sure this answer is correct. My friend asks me whether I can find k=16 algebraically. This is how I start it but can see a complication in my analysis. y = square root of [ 0.5*(x+1)^2 + k ] = > 4 
August 19th, 2012, 01:26 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,379 Thanks: 2011 
As 0.5*(x+1)² and ?(0.5*(x+1)² + k) are minimized when x = 1, put x = 1 to get ?k = 4, i.e. k = 16. That finds k by an algebraic method. 

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