# If a and b are positive numbers, prove that the equation a/x^3 + 2x^2 - 1  +  b/x^3 + x - 2  =  0 has at least one solution in the interval (-1,1)

Given that a>0, b>0 show that `a/(x^3+2x^2-1)+b/(x^3+x-2)=0` for at least one point on (-1,1).

(1) First note that `x^3+2x^2-1=(x+1)(x^2+x-1)` and `x^3+x-2=(x-1)(x^2+x+2)` . The denominator for the fraction with `a` is zero at `x=(-1+sqrt(5))/2` . Let `c=(-1+sqrt(5))/2` . Then on (c,1) `a/(x^3+2x^2-1)>0` and `b/(x^3+x-2)<0` .

(2) As x approaches c from the right, the fraction with numerator a grows without bound. (`x^3+2x^2-1 -> 0` and both numerator and denominator are positive.)

(3) As x approaches c from the right, the fraction with b in the numerator approaches -.873b.

(4) Thus no matter how large b is, eventually as x approches c from the right the first term will be larger than the second term in absolute value and the sum will be positive.

(5) As x approaches 1 from the left, the term with b in the numerator decreases without bound. (This term is always negative on (c,1), and the denominator is zero at 1). As x approaches 1 from the left, the term with a in the numerator approaches a/2. Then the second term will be greater in absolute value than the first term as x approaches 1 and the sum will be negative.

(6) The function is the sum of rational functions. Rational functions are continuous on their domains, and for both terms (c,1) is in the domain. Also, the sum of continuous functions is also continuous on the same interval.

(7) We have a continuous function where `f(x_1)>0` for some `x_1` near c, and `f(x_2)<0` for some `x_2` near 1. (`x_1` and `x_2` will depend on your choice of a and b). By the Intermediate Value Theorem, there exists `d in (c,1)` where f(d)=0.

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