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**lllll** Let

**a)** $\displaystyle f_1(x) = \min\{f_2(x), \ f_3(x)\}$, show that f_1(x) is continuous.

**b)** $\displaystyle f_4(x) = \max\{f_2(x), \ f_3(x)\}$ show that f_4(x) is continuous.

where $\displaystyle f_2(x)$ and $\displaystyle f_3(x)$ are both continuous.

**a)** I was thinking of splitting up into 2 where I assume firstly that $\displaystyle f_2(x)$ is the minimum secondly where $\displaystyle f_3(x)$. Thus if the minimum is $\displaystyle f_2(x)$ then $\displaystyle f_1(x) = f_2(x)$ thus for it to be continuous on any point c there exist a $\displaystyle \delta>0$ s.t. $\displaystyle 0<|x-c|<\delta \Rightarrow |f_1(x)-f_1(c)|= |f_2(x) - f_2(c)|<\varepsilon$.

now if I $\displaystyle f_3(x)$ is the minimum then just replace $\displaystyle f_2(x)$ by f_3(x).