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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