Prove Equation Has Solution in Given Interval

Calculus
Section 2.5

I will answer 4 Intermediate Value Theorem Questions later this afternoon. We end Section 2.5 today. As you can see, there's plenty of math to keep you busy for most of your Sunday. Enjoy. Be back later after spending a few hours at the Riegelmann Boardwalk at Coney Island or perhaps I'll go to Prospect Park.

 

given:
a>0
b>0

a/(x^3+2x^2-1)+b/(x^3+x-2)=0
has at least one solution inn interval (-1,1) => so, -1<x<1


Let us write

f(x)=x^3+2x^2-1 and g(x)=x^3+x-2

factor both

f(x)=(x + 1) (x^2 + x - 1)=(x+1)(x+(sqrt(5)+1)/2)(x-(sqrt(5)-1)/2)

g(x)=(x - 1) (x^2 + x + 2)=(x-1)((x+1/2)^2+7/4)


It appears that g(x)<0 all over the interval (-1,1) and it has a zero at x=1 and therefore for every positive number b the rational function b/g(x) is negative all over that interval and tends to -infinity as x→1- (from inside the interval (-1,1)).

On the other hand from the factorization of f(x) it is clear that f(x)>0 over the subinterval ((sqrt(5)-1)/2, 1) and it vanishes at its left end-point.

Hence for every positive number a the other rational function a/f(x) is positive over that subinterval and is tending to +infinity as x approaches the left end-point from inside that subinterval.

Putting all that together we conclude that the sum of these two rational functions is continuous over the subinterval

((sqrt(5)-1)/2, 1)

it is positive on the left part of that interval and turns to negative on the right part of it.
Hence, by the intermediate property of continuous function, that function must vanish at least at one point inside that subinterval.

 

It is amazing to me how you can easily answer any question I select from any textbook. You are truly a mathematical wizard.