We have the function $f(x)=\frac{4}{x^{2}+1}$ and the interval $[0,2]$. We can find the value of the function at $x=0$, $x=0.5$, $x=1$, and $x=2$.
At $x=0$, $f(0)=\frac{4}{0^{2}+1}=4$.
At $x=0.5$, $f(0.5)=\frac{4}{(0.5)^{2}+1}=\frac{16}{5}$.
At $x=1$,
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