SOLUTION 5: We are given the equation
$$ x^5-3x=4-x^2 \ \ \ \ \longrightarrow \ \ \ \ x^5+ x^2-3x-4=0 $$
Let function
$$ f(x)=x^5+ x^2-3x-4\ \ \ \ and \ choose \ \ \ \ m=0 $$
This function is continuous for all values of $x$ since it is a polynomial.
To establish an appropriate interval consider the graph of this function. (Please note that the graph of the function is not necessary for a valid proof, but the graph will help us understand how to use the Intermediate Value Theorem. On many subsequent problems, we will solve the problem without using the "luxury" of a graph.)
Note that $$ f(0)= (0)^{5}+(0)^2-3(0)-4 =-4<0 \ \ \ \ and \ \ \ \ f(2)= (2)^{5}+(2)^2-3(2)-4=26>0 $$
so that $$ f(0)=-4 < m <26=f(2) $$
i.e., $m=0$ is between $ f(0) $ and $ f(2) $. Now choose the interval to be $ \ [0, 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.,
$$ c^5+c^2-3c-4=0 $$
and the equation is solvable.
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