Q.Find local minimum value of the function $f$ given by $f(x) = 3 + |x|,\ x \in \mathbb{R}$.

The key idea is that $|x|$ has a sharp global minimum at $x = 0$, so the same point will give the local minimum of $f(x) = 3 + |x|$.

Step 1: Recall that $|x| \ge 0$ for all real $x$, and $|x| = 0$ only when $x = 0$.

Step 2: Therefore $f(x) = 3 + |x| \ge 3$, with equality exactly at $x = 0$. …

Concept and Intuition

A local minimum of a function $f$ at a point $c$ means that $f(c)$ is less than or equal to the function values at all points near $c$. In other words, there is some interval around $c$ where $f(c)$ is the smallest value.

For $f(x) = 3 + |x|$, the absolute value $|x|$ is always non-negative. It reaches its smallest possible value — exactly $0$ — only when $x = 0$. Since $|x| \ge 0$ for every real $x$, adding $3$ shifts the whole graph upward by $3$ units. So the smallest $f(x)$ can ever be is $3$, and that happens precisely at $x = 0$.

The graph is V-shaped, with the vertex at $(0, 3)$. On both sides of $x = 0$, the function increases (to the right, it rises with slope $1$; to the left, it rises with slope $-1$). So $x = 0$ is not just a local minimum — it is the global minimum.

Step-by-Step Reasoning

1. Understand the function's structure. $f(x) = 3 + |x|$ is the sum of a constant $3$ and the absolute value function. The absolute value is defined piecewise:

So $f$ can be written as:

2. Check the derivative (where it exists).

   For $x > 0$, $f'(x) = 1 > 0$ — the function is increasing.

   For $x < 0$, $f'(x) = -1 < 0$ — the function is decreasing.

   At $x = 0$, the derivative does not exist because the left-hand slope ($-1$) and right-hand slope ($1$) are different. This is the sharp corner of the V-shape.

3. Apply the definition of a local minimum.

   A point $c$ is a local minimum if there exists some $\delta > 0$ such that for all $x$ in $(c - \delta, c + \delta)$, $f(x) \ge f(c)$.

   Take $c = 0$. For any $x$ near $0$, $|x| \ge 0$, so $f(x) = 3 + |x| \ge 3 = f(0)$. …

Method: Finding Extrema of Functions Built From $|x|$

Functions containing an absolute value have a sharp corner where the inside expression is zero — a point where the usual "$f'(x)=0$" test cannot be used directly, since the derivative doesn't exist there.

Steps

Step 1: Rewrite the function piecewise

Use $|x-b| = x-b$ for $x \ge b$ and $|x-b| = -(x-b)$ for $x < b$ to split $f$ into two linear pieces meeting at $x=b$.

Step 2: Recognise the corner as a candidate extremum in its own right

At $x=b$, the left-hand slope and right-hand slope of the two pieces are generally different (here $-1$ and $+1$), so $f'(b)$ does not exist — but the definition of a local extremum only requires $f(c)\le f(x)$ (or $\ge$) nearby, not differentiability. So $x=b$ must still be checked directly.

Step 3: Confirm using the one-sided slopes …

Common Mistakes

Mistake 1: Concluding there is no minimum because $f'(0)$ doesn't exist

Why it's wrong: the definition of a local minimum only requires $f(c)\le f(x)$ for nearby $x$ — it does not require $f$ to be differentiable at $c$. The corner of $f(x)=3+|x|$ at $x=0$ is exactly such a non-differentiable point that is still a genuine (indeed global) minimum. Correct approach: when a function involves $|x|$, always check the corner point directly using the definition or one-sided slopes, rather than relying only on solving $f'(x)=0$.

Mistake 2: Trying to solve $f'(x)=0$ to locate the minimum …