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Quiz 2
Let $(M,d_1)=(N,d_2)=(\mathbb{R},d)$, where $d(x,y)=|x-y|$. Prove that $f \colon M \rightarrow N$ where $f(x)=5x^2-99x+7$ is continuous.
Proof: Let $\epsilon >0$ and let $p \in M$. Choose $0<\delta<\mathrm{min}\left\{ 1, \dfrac{\epsilon}{5+|10p-99|} \right\}$. Then if $p \in M$ with $|p-q|<\delta$, we compute
$$\begin{array}{ll}
|f(p)-f(q)|&=|(5p^2-99p+7)-(5q^2-99q+7)| \\
&= |5(p^2-q^2)-99(p-q)| \\
&=|p-q| \Big( |5(p+q)-99| \Big) \\
&=|p-q| \Big( |5(p+q-p+p)-99| \Big) \\
&=|p-q| \Big( |5(q-p+2p)-99| \Big) \\
&= |p-q| \Big( |5(p-q)+(10p-99)| \Big) \\
&\stackrel{\Delta\text{-inequality}}{\leq} |p-q|(5|p-q|+|10p-99|) \\
&\stackrel{|p-q|<\delta}{<} \delta(5\delta+|10p-99|) \\
&\stackrel{\delta<1}{<} \delta (5+|10p-99|) \\
&\stackrel{\delta<\frac{\epsilon}{5+|10p-99|}}{<} \dfrac{\epsilon}{5+|10p-99|} (5+|10p-99|) \\
&= \epsilon,
\end{array}$$
completing the proof. $\blacksquare$
