Extrema, Concavity, and Curve Analysis

Setting a derivative equal to zero produces candidates, not an optimization solution. A stationary point may be a maximum, minimum, or neither. An extreme value can also occur at a nondifferentiable point or an included endpoint. A systematic analysis keeps all of these possibilities visible.

Critical numbers and the first derivative test

A critical number is an interior input in the function’s domain where the derivative is zero or does not exist. An input excluded from the domain is not a critical number, although it can divide intervals of behavior. If $f$ is continuous at a critical number and $f'$ changes from positive to negative across it, $f$ has a local maximum there. Continuity at the critical number together with a negative-to-positive derivative change gives a local minimum. The conclusion follows from monotonicity on each side, established through MVT.

Continuity at the critical point cannot be omitted. Define $q(x)=-|x|$ for $x\ne0$ but set $q(0)=-1$. The derivative is positive immediately to the left and negative immediately to the right, yet $q(0)$ is a strict local minimum because nearby nonzero outputs are close to zero and greater than $-1$. Derivative signs control behavior on the neighboring intervals; continuity connects that behavior to the actual value at the join.

For $f(x)=x^3-3x$, $f'=3(x-1)(x+1)$. It is positive on $(-\infty,-1)$, negative on $(-1,1)$, and positive on $(1,\infty)$. Thus $f(-1)=2$ is a local maximum and $f(1)=-2$ a local minimum. Neither is an absolute extremum on all of $\mathbb R$, where the cubic is unbounded in both directions.

On the closed interval $[-2,3]$, EVT guarantees absolute extrema. Evaluate all interior critical points and both endpoints: $f(-2)=-2$, $f(-1)=2$, $f(1)=-2$, and $f(3)=18$. The absolute minimum is $-2$, attained at both $-2$ and $1$; the absolute maximum is $18$ at $3$. An algorithm that records only one minimizing input would lose a valid answer.

Concavity and second derivatives

Where $f'$ is increasing, the graph is concave up; where $f'$ is decreasing, it is concave down. A positive second derivative on an interval is sufficient for concavity up, and a negative second derivative for concavity down. An inflection point is a point on the graph where concavity changes across the point. The function must be continuous at the point under the convention used here. Merely solving $f''=0$ does not establish this change.

For the cubic above, $f''=6x$ changes sign at zero, so $(0,0)$ is an inflection point. For $x^4$, $f''=12x^2$ is zero at zero but positive on both sides. There is no inflection there. For $x^{1/3}$, concavity changes at zero even though the second derivative is undefined there; a point of the graph need not have a second derivative to be an inflection point under this definition.

The second derivative test says: if $f'(c)=0$ and $f''(c)>0$, a local minimum occurs; if $f''(c)<0$, a local maximum occurs, under differentiability in a neighborhood. When $f''(c)=0$, the test is inconclusive. The functions $x^4$, $-x^4$, and $x^3$ demonstrate all three possibilities at zero.

An organized curve investigation

Begin with the domain. Check symmetry and intercepts when useful. Investigate discontinuities and end behavior. Factor the first derivative to make a sign chart. Use the second derivative to inspect concavity. Evaluate important points and distinguish conclusions proved by signs from features merely suggested by a sketch.

For $g(x)=x+1/x$ on $x\ne0$, $g'=1-1/x^2$ vanishes at $\pm1$. It is negative for $0<|x|<1$ and positive for $|x|>1$. Thus $g(-1)=-2$ is a local maximum and $g(1)=2$ a local minimum. Its disconnected domain prevents treating those as global extrema: near zero the branches diverge, and their end behaviors are unbounded. Domain and asymptote checks come before any global claim.

Exercises

1. Analyze increasing and decreasing intervals of $x^3-6x^2+9x$.
2. Find its local extrema and inflection point.
3. Find the absolute extrema of $x^2-4x+1$ on $[0,5]$.
4. Explain why $x^4$ defeats the inference “zero second derivative means inflection.”
5. Find the absolute minimum of $|x-2|$ on $[0,3]$ and explain which derivative test requires care.
6. Explain why an excluded vertical-asymptote input is not a critical number.

Worked solutions

1. $f'=3(x-1)(x-3)$: increasing before $1$ and after $3$, decreasing between them.
2. Local maximum $f(1)=4$, local minimum $f(3)=0$. Since $f''=6x-12$ changes sign at $2$, the inflection point is $(2,2)$.
3. The stationary point is $2$ with value $-3$. Endpoint values are $1$ and $6$, giving absolute minimum $-3$ at $2$ and maximum $6$ at $5$.
4. Its second derivative is zero only at zero but concavity is up on both sides.
5. The minimum is zero at $2$, where the derivative does not exist. Setting only $f'=0$ would miss it.
6. A critical number must belong to the function’s domain. An asymptote still separates intervals to analyze.