SOLUTION 8: We are given the equation

$$ x^2-4x^3+1 = x-7 \ \ \ \ \longrightarrow \ \ \ \ x^2-4x^3-x+8=0 $$

Let function

$$ f(x)=x^2-4x^3-x+8 \ \ \ \ and \ choose \ \ \ \ m=0 $$

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

$$ f(0)= (0)^2-4(0)^3-(0)+8 = 8 > 0 \ \ \ \ and \ \ \ \ f(2)= (2)^2-4(2)^3-(2)+8 = -22 <0 $$

so that $$ f(2) = -22 < m < 8 = f(0) $$

i.e., $m=0$ is between $ f(0) $ and $ f(2) $.

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[0, 2]$ which satisfies $$ f(c)=m $$ i.e., $$ c^2-4c^3-c+8 =0 $$ and the equation is solvable.

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