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August 19th, 2012, 12:00 AM   #1
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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
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August 19th, 2012, 01:26 AM   #2
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As 0.5*(x+1) and ?(0.5*(x+1) + 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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