AP Calculus BC · Study guide

Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

AP Calculus · Analytical Applications of Differentiation

Extreme Value Theorem and Critical Points

Closed and bounded guarantees the extremes exist. Critical points and endpoints are where you look for them.

key idea$\text{Continuous on a CLOSED interval}\Rightarrow\text{both a max and a min exist}$

The theorem

The Extreme Value Theorem

Hypotheses — all must hold

$f$ is continuous on the interval.
The interval is closed and bounded: $[a,b]$, with both endpoints included.

What you may then conclude

$f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$. The theorem guarantees they exist; it does not say where.

Open the interval and the guarantee vanishes

$f(x)=x$ on the open interval $(0,2)$ is perfectly continuous, but it attains neither a maximum nor a minimum: for any candidate there is always a point closer to the excluded endpoints. The values $0$ and $2$ are approached but never reached.

closed interval → both extrema exist

open interval → no maximum

Left: continuous on $[-2,2]$ — both extrema exist, one at an interior critical point and one at an endpoint. Right: the same kind of function on an open interval, where the supremum is never attained.

Checking the hypotheses

Critical pointAn interior $c$ where $f'(c)=0$ or $f'(c)$ does not exist, with $f(c)$ defined. The only interior candidates for an extremum.

Local extremumLargest or smallest value compared with nearby points. Occurs only at critical points.

Absolute (global) extremumLargest or smallest value on the whole interval. Occurs at a critical point or at an endpoint.

Endpoint extremumAn absolute extremum occurring at $a$ or $b$. It is not a local extremum in the usual convention, because there is no neighbourhood on both sides.

Situation	Max guaranteed?	Min guaranteed?	Why
continuous on $[a,b]$	yes	yes	EVT applies
continuous on $(a,b)$	no	no	interval not closed
discontinuous on $[a,b]$	no	no	continuity fails
continuous on $[a,\infty)$	no	no	interval unbounded
continuous on $[a,b]$, no critical points	yes	yes	both at the endpoints

The last row is the one students doubt. With no critical points the function is monotonic, so the extremes sit at the two ends.

Worked examples

EX 1

Find every critical point

$f(x)=x^{3}-3x$

Find the critical points of $f(x)=x^{3}-3x$.

Differentiate

$f'(x)=3x^{2}-3=3\left(x^{2}-1\right)$.

Solve $f'=0$

$x^{2}=1$, so $x=\pm1$.

Look for where $f'$ fails to exist

$f'$ is a polynomial, defined everywhere, so there are none.

State them

Critical points at $x=-1$ and $x=1$.

Note what you have not shown

These are candidates. Nothing yet says either is a maximum or a minimum.

EX 2

A critical point where $f'$ is undefined

$f(x)=x^{2/3}$

Find the critical points of $f(x)=x^{2/3}$ on $[-1,8]$.

Differentiate

$f'(x)=\dfrac{2}{3x^{1/3}}$.

Solve $f'=0$

The numerator is the constant $2$, so $f'$ is never zero.

Find where $f'$ is undefined

At $x=0$, where the cube root vanishes.

Check $f$ is defined there

$f(0)=0$ exists, so $x=0$ qualifies as a critical point.

The lesson

Restricting attention to $f'=0$ would miss this entirely — and $x=0$ is where the absolute minimum sits.

EX 3

EVT guarantees, then you locate

$[-2,2]$

$f(x)=\tfrac13x^{3}-x+1$ on $[-2,2]$. Justify that extrema exist, then find them.

Justify existence

$f$ is a polynomial, so continuous on the closed bounded interval $[-2,2]$. By the EVT, absolute extrema exist.

Find the critical points

$f'(x)=x^{2}-1=0$ gives $x=\pm1$, both interior.

Build the candidate list

Critical points $x=-1,1$ plus endpoints $x=-2,2$.

Evaluate all four

$f(-2)=\tfrac13$, $f(-1)=\tfrac53$, $f(1)=\tfrac13$, $f(2)=\tfrac53$.

Select

Absolute maximum $\tfrac53$ at $x=-1$ and $x=2$; absolute minimum $\tfrac13$ at $x=-2$ and $x=1$. Extrema may occur at more than one point.

EX 4

When the guarantee is absent

open interval

Does $f(x)=x$ attain a maximum on $(0,2)$?

Check the hypotheses

$f$ is continuous, but $(0,2)$ is open — the second hypothesis fails.

Conclude about the theorem

The EVT gives no guarantee. That alone does not settle whether extrema exist.

Settle it directly

For any $x_{0}$ in $(0,2)$, the point $\tfrac{x_{0}+2}{2}$ is also in the interval and larger. So no maximum exists.

Name the values

The function approaches $2$ but never attains it — a supremum, not a maximum.

The distinction

“EVT does not apply” and “no extremum exists” are different claims. Here both happen to be true, but the second needs its own argument.

When it fails

Closed and bounded

$(a,b)$, $[a,\infty)$ and $(-\infty,\infty)$ all fail. Both endpoints must be present and finite.

A critical point is only a candidate

$f'(c)=0$ nominates $c$; it does not elect it. $f(x)=x^{3}$ has $f'(0)=0$ and no extremum there.

Do not forget $f'$ undefined

Cusps and corners are critical points too, and Example 2's minimum sits at exactly such a point.

Endpoints are candidates for absolute extrema

They cannot be local extrema in the usual convention, but they very often hold the absolute maximum or minimum.

Course map. Topic 5.2 · LO FUN-1.C — justify conclusions about the existence of extrema using the Extreme Value Theorem (EK FUN-1.C.1, FUN-1.C.2, FUN-1.C.3).