Graphs with prescribed extrema (x>0)

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For each item draw a different function, defined for $x>0$ and continuous (unless otherwise stated). The function must:

• (a) have both an absolute maximum and an absolute minimum;
• (b) have a relative maximum of value $3$, a relative minimum of value $1$, supremum $5$, and be unbounded below ($\inf=-\infty$);
• (c) have an absolute maximum, a finite infimum, and no relative minimum;
• (d) have an absolute maximum at the point where it also shows a jump discontinuity, and a non-absolute relative minimum.

Solution

These are construction tasks: it is enough to exhibit one valid example for each item. We give a model function and describe its features.

(a) We need a bounded function that actually attains both its greatest and its least value. Sine works: $\boxed{f(x)=\sin x,\qquad x>0}$ It is continuous, bounded between $-1$ and $1$, and attains both extremes infinitely often: absolute maximum $1$ (at $x=\tfrac\pi2,\ \tfrac{5\pi}2,\dots$) and absolute minimum $-1$ (at $x=\tfrac{3\pi}2,\dots$).

(b) No convenient elementary formula exists: we describe the graph. Near $x=0^+$ the curve drops to $-\infty$ (unbounded below, $\inf=-\infty$); then it rises to a relative maximum at height $3$; it falls to a relative minimum at height $1$; finally it rises again toward the horizontal asymptote $y=5$ without ever reaching it, so that $\sup=5$ (it is not a maximum).

(c) A function with a single hump: it rises, reaches the maximum, then decreases toward a finite value it never reaches. Model: $\boxed{f(x)=\dfrac{x}{1+x^2},\qquad x>0}$ It has absolute maximum $\tfrac12$ at $x=1$; as $x\to+\infty$ (and as $x\to0^+$) it tends to $0$, so $\inf=0$ is finite but not attained: there is no minimum (neither absolute nor relative).

(d) We build a piecewise function. A parabola with vertex (relative minimum) at height $1$ in $x=2$, a jump at $x=3$ where the value is $5$ (absolute maximum), then a decreasing branch that drops below $1$: $\boxed{f(x)=\begin{cases}(x-2)^2+1 & 0<x<3\\[2pt] 5 & x=3\\[2pt] \dfrac{3}{x-2} & x>3\end{cases}}$ At $x=3$ the left limit is $2$, the right limit is $3$: they differ, so this is a jump discontinuity; the value $f(3)=5$ is the absolute maximum. The vertex $(2;1)$ is a non-absolute relative minimum, because as $x\to+\infty$ the branch $\tfrac{3}{x-2}$ falls below $1$ (e.g. $f(6)=0.75$), with $\inf=0$.