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$f(x)=\max(2x+1,3-4x)$, where $x \in \mathbb{R}$. what is the minimum possible value of $f(x)$.

when, $2x+1=3-4x$, we have $x=\frac{1}{3}$

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Since $2x+1$ is strictly increasing and $3-4x$ is strictly decreasing, they must intersect at a some point, $z$. For any $\epsilon > 0, x =z+\epsilon$ implies that $2x+1 > 3-4x$ and similarly $x = z-\epsilon$ implies that $2x+1 < 3-4x$. Thus, the minimum of $f(x)$ must be at $z$. In your case, $z = \frac{1}{3}$.

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At $x = \frac13, f(x) = \frac53$ and this is the minimum value, since for $x>\frac13$, $2x+1>\frac53$ and for $x<\frac13$, $3-4x >\frac53$.

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The function $f$ is convex and its subdifferential is given by $$\partial f(x) = \begin{cases} \{-4\}, & x<\frac{1}{3}, \\ \, [-4,2], & x = \frac{1}{3}, \\ \{2\}, & x>\frac{1}{3}. \end{cases}$$Since $f$ is convex, then $\hat{x}$ minimizes $f$ iff $0 \in \partial f (\hat{x})$. It follows that the minimizing $\hat{x}$ is $\frac{1}{3}$, and hence the minimum value of $f$ is $\frac{5}{3}$.

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