Increasing, decreasing and concavity
“…if he will be content to begin with doubts, he shall end in certainties.”
Two different questions about a graph
A graph can be rising while becoming less steep, or falling while becoming less steep. The first derivative answers which direction the graph moves. The second derivative answers how its slopes change. Keeping those questions separate makes sign charts useful rather than mechanical.
Theorem
The sign of the first derivative
Let \(f\) be continuous on an interval \(I\) and differentiable at every interior point of \(I\). Then:
- \(f^{\prime}>0\) throughout implies \(f\) is strictly increasing.
- \(f^{\prime}<0\) throughout implies \(f\) is strictly decreasing.
- \(f^{\prime}=0\) throughout implies \(f\) is constant.
The non-strict conditions \(f^{\prime}\ge0\) and \(f^{\prime}\le0\) imply nondecreasing and nonincreasing behaviour, respectively.
Strictly increasing means \(f(u)<f(v)\) whenever \(u<v\). Nondecreasing allows equality: \(f(u)\le f(v)\).
The derivative conditions apply throughout the interior of \(I\). Continuity at any included endpoints extends the conclusion to those endpoints.
Proof (optional)
Take any \(u<v\) in the interval. The mean value theorem gives
\[f(v)-f(u)=f^{\prime}(c)(v-u)\]for some \(c\in(u,v)\). Since \(v-u>0\), the derivative sign gives the appropriate comparison of the values.
The strict derivative sign is sufficient, but is not necessary: \(x^3\) is strictly increasing even though its derivative is zero at one input.
Turning points from a sign change
Test
First derivative test
Suppose \(f\) is continuous at an interior critical number \(c\) and differentiable immediately to either side.
- A sign change \(+\to-\) in \(f^{\prime}\) gives a local maximum at \(c\).
- A sign change \(-\to+\) gives a local minimum.
- The same nonzero sign on both sides gives neither.
Continuity joins the increasing and decreasing pieces to the actual value at \(c\). The derivative at \(c\) itself need not exist.
Example 1
Make a sign chart, not a list of zeros
For \(f(x)=x^3-3x\), \(f^{\prime}(x)=3(x-1)(x+1)\). The critical numbers are \(-1\) and \(1\). Split the real line there and determine the sign on each interval:
| Input interval | \((-\infty,-1)\) | \((-1,1)\) | \((1,\infty)\) |
|---|---|---|---|
| Sign of \(x+1\) | \(-\) | \(+\) | \(+\) |
| Sign of \(x-1\) | \(-\) | \(-\) | \(+\) |
| Sign of \(f^{\prime}\) | \(+\) | \(-\) | \(+\) |
| Behaviour of \(f\) | increasing | decreasing | increasing |
Thus \((-1,2)\) is a local maximum and \((1,-2)\) is a local minimum. The derivative signs classify the inputs; substituting into \(f\) supplies their heights.
Concavity describes the change of slope
A differentiable function is concave up on an interval when \(f^{\prime}\) is nondecreasing, and concave down when \(f^{\prime}\) is nonincreasing. The graph bends toward increasing or decreasing tangent slopes, respectively. On an interval where the second derivative exists, the monotonicity theorem gives:
\[\begin{aligned} f^{\prime\prime}>0&\ \Longrightarrow\ \text{concave up},\\ f^{\prime\prime}<0&\ \Longrightarrow\ \text{concave down}. \end{aligned}\]For example, \(f(x)=e^{-x}\) has \(f^{\prime}<0\) and \(f^{\prime\prime}>0\). It falls, while its negative slopes become less steep: it is decreasing and concave up.
Definition
Inflection point
An inflection point is a point \((c,f(c))\) on a continuous graph where the concavity changes from one side to the other. Here we check concavity on intervals directly adjacent to \(c\).
A zero or undefined second derivative identifies a possible location; it does not establish a change. The input must belong to the graph.
Example 2
A change of concavity without a turn
For \(f(x)=x^3-3x\), the second derivative is \(6x\): negative for \(x<0\) and positive for \(x>0\). Hence \((0,0)\) is an inflection point. The slope there is \(f^{\prime}(0)=-3\), so the function is decreasing as it passes through the inflection.
An inflection need not have a horizontal tangent, and it need not be a local extremum.
A shortcut for a stationary point
Test
Second derivative test
Suppose \(f\) is twice continuously differentiable near \(c\) and \(f^{\prime}(c)=0\). Then:
- \(f^{\prime\prime}(c)>0\) gives a strict local minimum.
- \(f^{\prime\prime}(c)<0\) gives a strict local maximum.
- \(f^{\prime\prime}(c)=0\) leaves the test inconclusive.
Why the test works (optional)
If \(f^{\prime\prime}(c)>0\), continuity keeps \(f^{\prime\prime}\) positive on a small interval about \(c\). Thus \(f^{\prime}\) is strictly increasing there. Since \(f^{\prime}(c)=0\), its signs are negative to the left and positive to the right. The first derivative test gives a minimum. The negative case reverses the signs.
Example 3
The same inconclusive test, three outcomes
At zero, \(x^4\), \(-x^4\) and \(x^3\) all have first and second derivatives equal to zero. Yet they have, respectively, a local minimum, a local maximum, and neither.
Use the first derivatives \(4x^3\), \(-4x^3\) and \(3x^2\) to distinguish the cases. For \(x^4\) the second derivative \(12x^2\) is positive on both sides: its minimum is not an inflection. For \(x^3\) the second derivative \(6x\) changes sign: its flat crossing is an inflection.
Warning
Never infer an inflection from \(f^{\prime\prime}(c)=0\) alone. Never infer an extremum from \(f^{\prime}(c)=0\) alone. Split sign charts at missing domain inputs, but do not turn a missing point into an extremum or inflection.