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.