Can the minimum of a function on an interval [a,b ] be

Can the minimum of a function on an interval $\bigl[a,b\bigr]$ be greater than its maximum?

Step 1: Global extrema: no

By definition, if $m$ and $M$ are the least and greatest values of $f$ on $\bigl[a,b\bigr]$, then

\[ m \le f(x) \le M \qquad\text{for every } x\in\bigl[a,b\bigr], \]

and applying this at any single point gives $m\le M$. A global minimum can therefore never exceed a global maximum: the question answers itself.

Step 2: Local extrema: a different matter

$f$ has a local maximum at $x_{0}$ if $f(x_{0})\ge f(x)$ for all $x$ near $x_{0}$, and a local minimum at $x_{1}$ if $f(x_{1})\le f(x)$ for all $x$ near $x_{1}$. Each statement is confined to a neighbourhood, and nothing whatever links the values at two different points. So there is no reason why a local minimum value should not exceed a local maximum value — and it can.

Step 3: An explicit example

\[ f(x) = \begin{cases} -\bigl(x+1\bigr)^{2}+2, & -3\le x\le 0,\\[4pt] \bigl(x-2\bigr)^{2}+5, & 0\lt x\le 4. \end{cases} \]

On the first piece the vertex is at $x = -1$, giving a local maximum $f(-1) = 2$. On the second the vertex is at $x = 2$, giving a local minimum $f(2) = 5$. Hence

\[ \text{local minimum value } 5 \;\gt \; \text{local maximum value } 2. \]

Step 4: The graph

The second parabola sits bodily above the first, so its lowest point is higher than the first one's highest point.

Step 5: The same without a jump

A discontinuity is not needed — only enough room for the graph to climb between the two points. Take the smooth function

\[ g(x) = \frac{x}{2}+\cos x \quad\text{on}\quad \bigl[0,\,4\pi\bigr], \qquad g'(x) = \frac12-\sin x. \]

Here $g' = 0$ where $\sin x = \tfrac12$; the point $x = \dfrac{\pi}{6}$ is a local maximum and $x = \dfrac{5\pi}{6}+2\pi = \dfrac{17\pi}{6}$ is a local minimum, with

\[ g\!\left(\frac{\pi}{6}\right) = 1.1278248, \qquad g\!\left(\frac{17\pi}{6}\right) = 3.5845642 \;\gt \; 1.1278248. \]

The rising term $\dfrac{x}{2}$ lifts each successive oscillation above the previous one.

Step 6: Verification

\[f(-2) = 1 \lt f(-1) = 2 \gt f(0) = 1 \;\;\checkmark\ \text{(local maximum)}\]

\[f(1) = 6 \gt f(2) = 5 \lt f(3) = 6 \;\;\checkmark\ \text{(local minimum)}\]

\[\text{Globally on } \bigl[-3,4\bigr]:\ m = f(-3) = -2 \ \le\ M = f(4) = 9 \;\;\checkmark\]