x > -c > -b > -a, find the minimum value of (x+a)(x+b)/(x+c).?

Answer 1

quotient rule:

[(x+c)[(x+b)+(x+a)]-(x+a)(x+b)] / (x+c)^2
= [(x+c)(x+b) + (x+c)(x+a) - (x+a)(x+b)]/(x+c)^2
= [x^2+2cx+(cb+ca-ab)]/(x+c)^2.

Using the quadratic formula on the numerator:

x=-c +/- sqrt((c-a)(c-b))

Since x > -c, choose the + sign to obtain the critical point. It can be substituted into the original function to obtain the minimum.

Note: it should be verified that this really is a minimum, not a maximum, by using the second derivative test or examining the behavior of the first derivative near the critical point to see if it changes from + to - (maximum) or - to + (minimum).

Answer 2

ok. x>-c>-b>-a. which is the highest negative number = -1.

(x+a)(x+b) = x^2+(a+b)x+ab. differentiating 2x+a+b.

x+c. differentiating (x+c) = 1.

Now 2x+a+b/1 is the curvature of the curve which for minima > 0.

so again differentiating 2x+a+b = 2 > 0 so its a minima.

x = -(a+b)/2 will be the minima. Substituting in the equation.

(a-a/2-b/2)(b-b/2-a/2)/(c-(a+b)/2) = (a/2-b/2) (b/2-a/2)/......

= -(a-b)^2/2(2c-a-b) is the min. value......;-)

Answer 3

Minus infinity...if all constants (c, b, a) = 0 then you have (x^2/x) = x = ? Which means, when all the constants are zero, you can set x to anything you want to including -infinity, which is just about as minimum one can get.

Answer 4

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