Reading a Function’s Shape from Two Derivatives
A first-derivative sign chart describes whether a function rises or falls. A second-derivative sign chart describes how its tangent slopes change, and therefore whether the graph is concave up or concave down. Combining these descriptions gives a more informative picture than either one alone.
The information must also agree. For example, if \(f''>0\) throughout an interval, then \(f'\) increases there. A chart claiming that \(f'\) is positive at one point and later negative at another point on that same interval cannot be correct. Checking such consistency is useful when analyzing a function and when interpreting derivative tables or verbal descriptions.
What Each Sign Tells Us
On an interval where \(f'(x)>0\), \(f\) is increasing; where \(f'(x)<0\), \(f\) is decreasing. Where \(f''(x)>0\), the graph is concave up; where \(f''(x)<0\), it is concave down. These interval conclusions follow from the derivative tests for monotonicity and concavity.
At an interior point \(c\), a change in \(f'\) from positive to negative indicates a local maximum, and a change from negative to positive indicates a local minimum, provided \(f\) is continuous at \(c\) and differentiable on the intervals immediately to its left and right. If \(f\) is differentiable throughout a neighborhood of \(c\), continuity at \(c\) is automatic. A derivative sign that does not change does not establish a local extremum.
- Increasing and concave up: the graph rises while its tangent slopes increase.
- Increasing and concave down: the graph rises while its tangent slopes decrease.
- Decreasing and concave up: the graph falls while its tangent slopes increase, becoming less negative if they remain negative.
- Decreasing and concave down: the graph falls while its tangent slopes decrease, becoming more negative if they remain negative.
These combinations describe behavior over intervals, not necessarily at a single point. At a boundary between intervals, the function may be continuous while its derivative is undefined or changes abruptly. In that case, analyze each side and check the function’s value at the boundary separately.
Why Concavity Restricts the First Derivative
If \(f''>0\) throughout an interval, then \(f'\) is strictly increasing there. To justify this, take any two points \(a<b\) in the interval. The function \(f'\) is continuous on \([a,b]\) and differentiable on \((a,b)\), since \(f''\) exists there. The Mean Value Theorem applied to \(f'\) gives a point \(t\in(a,b)\) such that
Both \(f''(t)>0\) and \(b-a>0\), so \(f'(b)-f'(a)>0\). Thus \(f'(b)>f'(a)\). The same reasoning shows that if \(f''<0\) throughout an interval, then \(f'\) is strictly decreasing there.
The points \(a\) and \(b\) in this argument are chosen inside the interval. This matters: applying the Mean Value Theorem on a closed interval that includes a boundary requires the relevant function to be continuous at that boundary. Without that hypothesis, the theorem cannot be used there.
Extrema and Concavity Near a Point
When \(f\) is twice differentiable near an interior point \(c\), a strict local minimum is consistent with concavity up near \(c\), and a strict local maximum is consistent with concavity down near \(c\). Here is a justification for the concave-up case.
If \(f\) has a local extremum at the interior point \(c\) and is differentiable there, Fermat’s theorem gives \(f'(c)=0\).
If \(f''>0\) on a neighborhood of \(c\), the preceding argument shows that \(f'\) is strictly increasing there. Therefore \(f'(x)<0\) for \(x<c\) nearby and \(f'(x)>0\) for \(x>c\) nearby.
The function decreases to \(c\) from the left and increases from \(c\) to the right. Hence \(c\) is a local minimum, not a local maximum.
Reversing the signs gives the corresponding result for \(f''<0\): a stationary point in a neighborhood of concavity down is a local maximum. These conclusions require the stated differentiability and concavity conditions. A positive second derivative at just one point, without control on a neighborhood, is not enough to establish the shape there.
Worked Example: Combining a Cubic’s Increase, Decrease, and Concavity
Worked Example: A Local Maximum and a Local Minimum
Analyze the intervals of increase and decrease, the concavity, and the local extrema of \(f(x)=x^3-3x\).
Differentiate once:
The critical numbers are \(x=-1\) and \(x=1\). The factors show that \(f'(x)>0\) on \((-\infty,-1)\), \(f'(x)<0\) on \((-1,1)\), and \(f'(x)>0\) on \((1,\infty)\). Thus \(f\) increases, then decreases, then increases. Since this polynomial is continuous, the positive-to-negative change at \(-1\) gives a local maximum, while the negative-to-positive change at \(1\) gives a local minimum.
Calculate their function values:
Therefore the local maximum is \((-1,2)\), and the local minimum is \((1,-2)\).
Now differentiate again:
For \(x<0\), \(f''(x)<0\), so the graph is concave down. For \(x>0\), \(f''(x)>0\), so it is concave up. The maximum at \(-1\) lies in a region of concavity down, and the minimum at \(1\) lies in a region of concavity up. The local behavior is consistent with both derivative descriptions.
Check: At \(x=-1\), \(f'(-1)=3(1)-3=0\), and at \(x=1\), \(f'(1)=3(1)-3=0\), as expected at these differentiable local extrema. The computed function values are obtained by direct substitution into \(f\).
Worked Example: A Quartic with Several Shape Changes
Worked Example: Matching Four Monotonicity Intervals to Concavity
For \(g(x)=\frac{x^4}{4}-\frac{x^2}{2}\), find the increasing and decreasing intervals, the concavity intervals, and the local extrema.
First compute:
The critical numbers are \(-1\), \(0\), and \(1\). Checking the factors gives \(g'<0\) on \((-\infty,-1)\), \(g'>0\) on \((-1,0)\), \(g'<0\) on \((0,1)\), and \(g'>0\) on \((1,\infty)\). Hence \(g\) decreases, increases, decreases, and increases on those consecutive intervals. The sign changes give local minima at \(x=-1\) and \(x=1\), and a local maximum at \(x=0\).
Substitution gives:
Thus the local minima are \((-1,-\frac14)\) and \((1,-\frac14)\), and the local maximum is \((0,0)\).
For concavity, compute:
Solving \(g''(x)=0\) gives \(3x^2=1\), so \(x=\pm\frac{1}{\sqrt{3}}\). Since \(3x^2-1>0\) when \(|x|>\frac{1}{\sqrt{3}}\), \(g\) is concave up on \((-\infty,-\frac{1}{\sqrt{3}})\) and \((\frac{1}{\sqrt{3}},\infty)\). It is concave down on \((-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})\).
The minima at \(-1\) and \(1\) lie in concave-up intervals, and the maximum at \(0\) lies in a concave-down interval. This is consistent with the local behavior. Also, the signs of \(g'\) agree with the direction of change on each interval; for example, on \((0,1)\) the function decreases even though the concavity changes at \(\frac{1}{\sqrt{3}}\). Concavity changing does not by itself force the function to switch from increasing to decreasing.
Check: \(g'(-1)=(-1)^3-(-1)=0\), \(g'(0)=0\), and \(g'(1)=1-1=0\). The second-derivative values at the extrema are \(g''(-1)=2\), \(g''(0)=-1\), and \(g''(1)=2\), matching the concavity descriptions.
Worked Example: Testing Whether Derivative Information Can Coexist
Worked Example: Detecting an Impossible Sign Chart
Suppose a report claims that \(f''(x)>0\) throughout \((0,2)\), while also stating that \(f'(\frac12)=2\) and \(f'(\frac32)=-1\). Can all three claims be true?
Because \(f''>0\) on \((0,2)\), the derivative \(f'\) is strictly increasing there. Since \(\frac12<\frac32\), strict increase requires
The reported values would instead require \(2<-1\), which is false. Therefore, the claims cannot all describe the same function.
The Mean Value Theorem gives the contradiction directly. Applied to \(f'\) on \([\frac12,\frac32]\), it gives some \(t\in(\frac12,\frac32)\) such that
But the left side from the reported values is \(-1-2=-3\), not positive. The error must be in at least one claim: the concavity, one of the derivative values, or the assertion that they refer to the same interval.
Check: The interval used for the Mean Value Theorem lies wholly inside \((0,2)\), so \(f'\) is continuous on the closed interval and differentiable in its interior. The theorem’s conditions are satisfied.
A Pitfall: A Derivative Sign Change Needs Continuity at the Point
It is tempting to apply the first-derivative sign test using only the signs on either side of a point. Continuity at the point is an essential condition. For example, define
For \(x<0\), \(f'(x)=1>0\); for \(x>0\), \(f'(x)=-1<0\). A mistaken application would say “positive to negative, therefore a local maximum at \(0\).” But \(f(0)=-1\), while for every nearby \(x\ne0\), \(f(x)=-|x|\), and \(-|x|>-1\) whenever \(0<|x|<1\). Thus \(f(x)>f(0)\) on both sides close to \(0\), and \(0\) is a local minimum. The function is not continuous at \(0\), so the sign-test conclusion is not justified.
With continuity at \(c\), the sign test follows from the Mean Value Theorem. If \(f'>0\) to the left of \(c\), then for \(x<c\) sufficiently close to \(c\), the theorem applied to \(f\) on \([x,c]\) gives \(f(c)-f(x)=f'(t)(c-x)>0\) for some \(t\in(x,c)\). Thus \(f(x)<f(c)\). If \(f'<0\) to the right, applying the theorem on \([c,x]\) gives \(f(x)-f(c)=f'(t)(x-c)<0\). Together these inequalities establish a local maximum. The local-minimum case follows by reversing the derivative signs.
Using the Combined Picture
When interpreting derivative information, keep the claims attached to their intervals and points. Concavity constrains how \(f'\) can change within one interval, but a boundary can have different behavior from either neighboring interval. An interior extremum of a differentiable function must have derivative zero; a sign change in \(f'\), when the continuity conditions hold, identifies the direction of the extremum. Concavity then provides an additional consistency check.
A reliable analysis therefore asks not only “Where is the function increasing?” and “Where is it concave up?” but also “Could those statements both be true?” and “Do they support the claimed local extrema?” That combined reading helps distinguish a coherent description of a graph from a sign chart with conflicting information.
Check Your Understanding
Use derivative signs and concavity together. State any conditions needed for your conclusions.
- If \(f''>0\) on an interval, can \(f'\) be positive at an earlier point and negative at a later point in that interval? Explain.
- A continuous function has \(f'<0\) immediately left of \(c\) and \(f'>0\) immediately right of \(c\). What local-extremum conclusion follows?
- For \(f''<0\) near a differentiable stationary point, what type of local extremum is consistent with the concavity?
- Why does a concavity change not necessarily mean that the function changes from increasing to decreasing?
- What condition is missing if a derivative-sign change is used to classify an extremum at a point where the function is not known to be continuous?
