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When x is real what is the least value of
(x**2-6*x+5)/(x**2+2*x+1)
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Answer Posted / kermit rose
(x**2 - 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it's numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 - (x-1)(x-5)( 2 (x+1) )
= (2 x - 6) (x+1)**2 - (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) - (x-1)(x-5) = 0
(x**2 - 2 x - 3 ) - (x**2 - 6 x + 5) = 0
x**2 - x**2 - 2 x + 6 x - 3 - 5 = 0
4 x - 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 - 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 - 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
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