SOLUTION 1: We are given the equation $ 3x^5-4x^2=3 $ and the interval $ [0, 2] $. Let function
$$ f(x)=3x^5-4x^2 \ \ \ \ and \ choose \ \ \ \ m=3 $$
This function is continuous for all values of $x$ since it is a polynomial. (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 $$ f(0)= 3(0)^5-4(0)^2= 2<3 \ \ \ \ and \ \ \ \ f(2)= 3(2)^5-4(2)^2=80>3 $$
so that $$ f(0)=2 < m < 80=f(2) $$
i.e., $m=3$ 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
$$ f(c)=m $$
$$ 3c^5-4c^2= 3 $$
and the equation is solvable.
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