SOLUTION 12: We are given the equation

$$ \displaystyle{ x-1 \over x^2+2 } = \displaystyle{ 3-x \over x+1 } \ \ \ \ \longrightarrow \ \ \ \ (x-1)(x+1)=(x^2+2)(3-x) $$ $$ \longrightarrow \ \ \ \ x^2+x-x-1 = 3x^2-x^3+6-2x $$ $$ \longrightarrow \ \ \ \ x^3-2x^2+2x-7=0 $$

Let function

$$ f(x)= x^3-2x^2+2x-7 \ \ \ \ and \ choose \ \ \ \ m=0 $$

This function is continuous for all values of $x$ since it is a polynomial. We now need to search for an appropriate interval satisfying the assumptions of the Intermediate Value Theorem. By trial and error, we have that

$$ f(0)= (0)^3-2(0)^2+2(0)-7 = -7 < 0 \ \ \ \ and \ \ \ \ f(3)= (3)^3-2(3)^2+2(3)-7 = 8 > 0 $$

so that $$ f(0) = -7 < m < 8 = f(3) $$

i.e., $m=0$ is between $ f(0) $ and $ f(3) $.

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[0, 3]$ which satisfies $$ f(c)=m $$ i.e., $$ c^3-2c^2+2c-7 =0 $$ and the equation is solvable.

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