SOLUTION 16: Consider the graphs of $y=f(x)$ and $y=x^2$ on the interval $ [0, 2]$ with $ f(0)>0 $ and $ f(2)<4 $.

Let function $$ g(x)= f(x)-x^2 \ \ \ \ and \ choose \ \ \ \ m=0 $$

Function $g$ is continuous on the interval $ [0, 2] $ since it is the DIFFERENCE of continuous functions. Note that $$ g(0) = f(0) - (0)^2 = f(0)>0 $$ and $$ g(2)=f(2) - (2)^2 = f(2)-4 < 0 $$

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

The assumptions of the Intermediate Value Theorem have now been met on the interval $[0, 2]$, so we can conclude that there is some number $c$ in the interval $[0, 2]$ which satisfies $$ g(c)=m $$ i.e., $$ f(c)-c^2 =0 \ \ \ \ \longrightarrow \ \ \ \ f(c)=c^2 $$ This completes the proof.

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