Given:
f(x) = √(x^2-10x+314) + √(x^2+20x+325)
Determine the minimum value of |f(x)| for a real number x.
*** Adapted from a problem appearing in 2017 Singapore M.O. open.
Note that if you find the minimum of each term separately, the minimum of the first term is √289 (at x=5) and √225 (at x= -10).
So the min of f(x) is >= 17+15 = 32. Which doesn't really help. But was a little bit fun.
Take derivative and set equal to zero.
f'(x) = ((2x-10)/2√(x^2-10x+314)) + ((2x+20)/2√(x^2+20x+325))
f'(x) numerator = (x-5)√(x^2+20x+325) + (x+10)√(x^2-10x+314)
f'(x) denominator = √(x^2-10x+314) * √(x^2-10x+314)
Set f'(x) numerator to zero
(x-5)√(x^2+20x+325) + (x+10)√(x^2-10x+314) = 0
(x-5)√(x^2+20x+325) = -(x+10)√(x^2-10x+314)
square both sides
(x^2 - 10x +25)(x^2+20x+325) = (x^2 + 20x + 100)(x^2-10x+314)
x^4 + 10x^3 + 150x^2 - 2750x + 8125 = x^4 + 10x^3 + 214x^2 + 5280x + 31400
150x^2 - 2750x + 8125 = 214x^2 + 5280x + 31400
64x^2 + 8030x + 23275 = 0
which has roots of
-95/32 = -2.96875 and
-245/2 = -122.5
f(-95/32) = 35.34119409414458 <-- this is the minimum
f(-245/2) = 242.12393520674487
https://www.desmos.com/calculator/kx0v6dp0us
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Posted by Larry
on 2024-07-10 23:59:43 |
Solution
