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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. $\blacksquare$