SOLUTION 14: We are given the equation $$ x^3 = 2x^2+3x-3 \ \ \ \ \longrightarrow \ \ \ \ x^3-2x^2-3x+3=0 $$ Let function $$ f(x)=x^3-2x^2-3x+3 \ \ \ \ and \ choose \ \ \ \ m=0 $$ This function is continuous for all values of $x$ since it is a polynomial. We know from algebra that a cubic polynomial has at most three distinct real roots. To establish three distinct and appropriate intervals, consider the graph of this function.

INTERVAL ONE: Note that $$ f(0)= (0)^3-2(0)^2-3(0)+3 =3<0 \ \ \ \ and \ \ \ \ f(-1)= (-2)^3-2(-2)^2-3(-2)+3 = -7<0 $$

so that $$ f(-2) = -7 < m < 3=f(0) $$

i.e., $m=0$ is between $ f(-2) $ and $ f(0) $. Now choose the interval to be $ \ [-2, 0] $.

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

INTERVAL TWO: Note that $$ f(0)= (0)^3-2(0)^2-3(0)+3 =3<0 \ \ \ \ and \ \ \ \ f(1)= (1)^3-2(1)^2-3(1)+3 = -1<0 $$

so that $$ f(1) = -1 < m < 3=f(0) $$

i.e., $m=0$ is between $ f(0) $ and $ f(1) $. Now choose the interval to be $ \ [0, 1] $.

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[0, 1]$ which satisfies $$ f(c)=m $$ i.e., $$ c^3-2c^2-3c+3 =0 $$ and the equation is solvable on this interval.

INTERVAL THREE: Note that $$ f(3)= (3)^3-2(3)^2-3(3)+3 = 3>0 \ \ \ \ and \ \ \ \ f(1)= (1)^3-2(1)^2-3(1)+3 = -1<0 $$

so that $$ f(1) = -1 < m < 3=f(3) $$

i.e., $m=0$ is between $ f(1) $ and $ f(3) $. Now choose the interval to be $ \ [1, 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 $[1, 3]$ which satisfies $$ f(c)=m $$ i.e., $$ c^3-2c^2-3c+3 =0 $$ and the equation is solvable on this interval. This completes the solution.

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