Consider the function:
f(x) = √(x-1) + √(x+1) - √(2x)
So the two expressions are f(6) and f(7).
Note that f(x) is defined only for x>=1
We just need to know if the derivative of f(x) is positive or negative near, say, x= 6.5
f'(x) = 1/2√(x-1) + 1/2√(x+1) - 1/√(2x)
f'(x) = Numerator / Denominator
Numerator = √(2x)[√(x+1) + √(x-1)] - 2√((x+1)(x-1))
Denominator = 2√(2x)√((x+1)(x-1))
Now approximate √(2x)[√(x+1) + √(x-1)] with 2√2 * x
and approximate √((x+1)(x-1)) with 2 * x
Now the approximate Numerator = 2√2 * x - 2x = 2(√2 - 1)x which is clearly positive.
The approximations made the left positive term become more positive but also made the negative left term more negative. Furthermore, the magnitude of the error is about 4 times larger for the negative term, so the approximation gives a result that is more negative than the exact value. And yet it is still positive.
So f(7) > f(6) which is the answer being sought.
note that if you graph both f'(x) and the approximation of f'(x)~ (√2 - 1)/√(2x), as x get greater than about 2, the two curves are very close, the approximation being a tiny bit smaller.
Posted by Larry
on 2018-02-23 10:28:06