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

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this is problem number seventy of the Stuart Calculus eighth edition section two point five. To pay A and B are positive numbers. Prove that equation a divided by the quantity X cube two plus two X squared minus one plus Be ready by the quantity X cubed plus X minus two is equal to zero. Prove that this equation has at least one solution. Interval negative one to one. So we can, ah, work with dysfunction as is, or we can even modify a little bit to make our calculation a bit easier. So instead of calling this whole upside at Lex, we make a slight modification to eliminate the factions and multiply by this denominator executed close to X squared and this one and by the other denominator x cubed plus X minus two. And if we want to play this to both sides, we get that we get eight times X cubed plus X minus two plus B times execute plus two x squared like this one, and this equals zero. So just for our comedians on this is what we have chosen to be our function f. So we're goingto prove that there's at least one solution to this function equally. Zero. Um, basically, there is a root that exists between the interruption. I get a one to one and we're gonna evaluate negative f of negative one and another one to prove this. That's a negative one. Eyes eight times Negative. One minus one minus two. Plus be not quite O'Brien. Negative. One plus two on this one. And if we simplify this like the one plus two minus one, this will be zero and a one. A two run minus twos. NATO forces will be needed for a for half of the negative one and forever one. We should have a more compliant buying. One plus one makes two. It's being times the quantity one plus two minus one. And this should be equal to just to be here. And since we were given initially that A and B are positive numbers, this is a negative number. Weston zero. This is a positive number, Cris ta Modified. This function is polynomial. Then according to the intermediate value there, um thie function dysfunction on the central takes all the values from negative for eight to be including zero, and it definitely takes this value within this interval from negative once one. So we have confirmed that this function as his taste has a root, and this again is equivalent to the initial equation. And so this equation has at least one solution in the interval from negative one to one.

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