Connecting f, f′, f″ and the Extreme Value Theorem

Extreme Value Theorem

If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$. Both hypotheses matter: $f(x)=\dfrac1x$ on $(0,1]$ has no maximum (not closed), and a function with a jump can skip its extreme value (not continuous). EVT is why the candidates test works — the extremes exist and can only sit at critical points or endpoints.

On an open interval EVT is silent. If $c$ is the only critical point and $f'$ changes sign there, the local extremum is also the absolute one: "Since $x=c$ is the only critical point and $f'$ changes from negative to positive at $c$, $f(c)$ is the absolute minimum."

One graph, three readings

Given the graph of $f'$: where $f'\gt0$, $f$ increases; where $f'$ is increasing, $f''\gt0$ and $f$ is concave up; where $f'$ has a local extremum, $f$ has a point of inflection. Given the graph of $f$: concave up means the slopes increase, so $f'$ is increasing and $f''\gt0$.

Worked example: $f(x)=\dfrac{x^3}{3}-2x^2+3$ on $[0,5]$, so $f'(x)=x^2-4x=x(x-4)$ and $f''(x)=2x-4$. $f'\lt0$ on $(0,4)$, $f'\gt0$ on $(4,5)$ ⇒ minimum at $x=4$, $f(4)=-\dfrac{23}{3}$. Candidates: $f(0)=3$, $f(5)=-\dfrac{16}{3}$ ⇒ absolute maximum $3$ at $x=0$. $f''$ changes sign at $x=2$ ⇒ inflection point at $x=2$, exactly where the graph of $f'$ bottoms out.