Introduction
A stock records how much of something is present at a particular time: for example, the number of items in inventory or the amount of water in a reservoir. A turning point in a stock tells us when it changes direction, from rising to falling or from falling to rising. To locate such points, we examine the sign of the stock’s derivative.
A cumulative quantity is different: it records an amount accumulated over time. Its turning points depend on the rate at which that cumulative quantity changes, which need not be the same function as the stock itself. Keeping the quantity and its rate distinct is essential for interpreting a model correctly.
Local Extrema and Derivative Signs
A stock \(S(t)\) is increasing where \(S'(t)>0\) and decreasing where \(S'(t)<0\). Therefore, a derivative changing from positive to negative at \(c\) guarantees a local maximum there; a change from negative to positive guarantees a local minimum. These are sufficient tests: when their conditions hold, the claimed extremum follows.
The reason is that a positive derivative makes the function increase on the left interval, and a negative derivative makes it decrease on the right interval. The Mean Value Theorem justifies each monotonicity claim: for two points in either interval, the difference in function values equals the derivative at an intermediate point multiplied by the difference in inputs. Continuity at \(c\) then gives the comparisons with \(f(c)\).
The sign-change tests are not necessary. A function can have a local extremum even if its derivative does not keep one sign on each side of that point. A different necessary condition applies when the function is differentiable at an interior extremum.
To see why the derivative must vanish, suppose first that \(c\) is a local maximum. For sufficiently small positive \(h\), \(f(c+h)-f(c)\le 0\), so the right difference quotient is nonpositive. For sufficiently small negative \(h\), \(f(c+h)-f(c)\le 0\), but division by \(h<0\) makes the left difference quotient nonnegative. If the derivative exists, the two one-sided limits agree. Their common value must therefore be both nonpositive and nonnegative, hence zero. At a local minimum the inequalities reverse, and the same conclusion follows.
Worked Example: A Stock Reaches a Local Maximum
Worked Example: Finding a Turning Point in Inventory
Suppose inventory, measured in items, is modeled for \(0\le t\le 5\) hours by \(S(t)=100+12t-3t^2\). Find and classify its interior turning point.
Differentiate the stock:
Set the derivative equal to zero to find a stationary point:
For \(0\le t<2\), \(12-6t>0\), so the stock is increasing. For \(2<t\le5\), \(12-6t<0\), so the stock is decreasing. The derivative changes from positive to negative, which gives a local maximum at \(t=2\). Its value is
Check by substitution into the derivative: \(S'(2)=12-6(2)=0\). Also, for every \(t\), \(S(t)=112-3(t-2)^2\), so \(S(t)\le112\), with equality at \(t=2\). This confirms the maximum directly.
Worked Example: A Cumulative Total Turns at Its Own Rate’s Sign Change
Worked Example: Classifying Extrema of a Cumulative Quantity
Let a cumulative quantity \(C(t)\), measured in units, have rate \(C'(t)=q(t)=(t-1)(t-3)\) on \(0\le t\le4\). Find and classify its interior turning points. The constant initial value of \(C\) is irrelevant to their locations.
The rate is zero at \(t=1\) and \(t=3\). Check its sign on the intervals between and around those times:
Since the factors change sign only at \(1\) and \(3\), \(q\) is positive before \(1\), negative between \(1\) and \(3\), and positive after \(3\). Thus \(C\) changes from increasing to decreasing at \(1\), giving a local maximum, and from decreasing to increasing at \(3\), giving a local minimum.
For a direct check, integrate the rate:
where \(K\) is a constant. Differentiating gives \(C'(t)=t^2-4t+3=(t-1)(t-3)\), as specified. The critical values are
The first value is a local maximum and the second a local minimum. The additive constant shifts the cumulative total vertically but does not change its rate or the turning-point times.
Why a Stock’s Turning Point Need Not Be a Cumulative Turning Point
For any differentiable stock \(S(t)\), a turning point of the stock is about the sign of \(S'(t)\): the stock’s rate of change. For a cumulative quantity \(C(t)\), its turning points are about the sign of \(C'(t)\), its own rate. These rates are not automatically the same.
For example, suppose a cumulative total is formed by accumulating the stock itself, so that \(C'(t)=S(t)\). If the stock remains positive, then \(C\) is increasing, even when the stock has a local maximum. In the inventory example, \(S(t)=112-3(t-2)^2\) is positive throughout \(0\le t\le5\). Consequently, \(C'(t)=S(t)>0\) throughout that interval, so the cumulative total has no interior turning point there. The stock peaks at \(t=2\), but the cumulative total does not.
In another model, a cumulative total might accumulate a flow \(q(t)\), so \(C'(t)=q(t)\). Then the cumulative total turns when the flow changes sign, not merely when some related stock has a maximum or minimum. Always identify the quantity being differentiated before applying a sign test.
Worked Example: An Extremum Without a Derivative Sign Change
Worked Example: A Strict Maximum with Oscillating Derivative
Define \(f(0)=0\) and, for \(x\ne0\), let \(f(x)=-x^2(2+\sin(1/x))\). Show that \(f\) has a strict local maximum at zero even though its derivative takes both signs arbitrarily close to zero.
Since \(-1\le\sin(1/x)\le1\), we have \(1\le2+\sin(1/x)\le3\). Thus, for every \(x\ne0\),
This proves that zero is a strict local maximum; in fact, the inequality holds for every nonzero \(x\).
First check differentiability at zero using the difference quotient:
Its absolute value is at most \(3|h|\), which tends to zero as \(h\to0\). Therefore \(f'(0)=0\). For \(x\ne0\), the product and chain rules give
To verify that both signs occur arbitrarily close to zero, take \(x_n=1/(2\pi n)\) for positive integers \(n\). Then \(\sin(1/x_n)=0\), \(\cos(1/x_n)=1\), and \(f'(x_n)=1-4x_n\), which is positive for \(n\ge1\). Next take \(y_n=1/((2n+1)\pi)\). Then \(\sin(1/y_n)=0\), \(\cos(1/y_n)=-1\), and \(f'(y_n)=-1-4y_n<0\). Both sequences approach zero. Hence the derivative does not have a fixed positive sign on one side or a fixed negative sign there, despite the strict maximum.
This example confirms the distinction: the derivative sign-change test is sufficient, not necessary. Fermat’s theorem still applies, because \(f'(0)=0\).
Common Pitfalls
Pitfall 1: Treating a zero derivative as proof of a turning point. For \(g(t)=t^3\), \(g'(t)=3t^2\), so \(g'(0)=0\). But \(g(t)<0\) for \(t<0\) and \(g(t)>0\) for \(t>0\); zero is not a local maximum or minimum. A stationary point may fail to be an extremum.
Correction: Use the derivative’s signs on both sides to apply a sufficient sign-change test, or compare function values directly. Do not infer an extremum from \(f'(c)=0\) alone. Conversely, do not insist that a derivative must change sign in order for an extremum to exist; the oscillating example demonstrates why.
Pitfall 2: Using the stock’s rate to classify a cumulative total without checking what the total accumulates. The correct test for a cumulative quantity \(C\) uses \(C'\). If \(C'=S\), the relevant sign is the stock’s sign; if \(C'=q\), it is the flow’s sign. First write the model relationship, then analyze the appropriate derivative.
Finally, these tests concern interior points. At an endpoint of a time interval, a model may have a one-sided maximum or minimum even when the derivative does not vanish there. Fermat’s theorem requires an interior point, so it does not rule out endpoint extrema.
Check Your Understanding
Use the definitions and derivative-sign reasoning above to answer the following questions.
- If a stock’s derivative is positive just before \(t=c\) and negative just after, what kind of local extremum does the stock have at \(c\)?
- What does Fermat’s theorem say about the derivative at a differentiable interior local minimum? Does that condition alone guarantee a minimum?
- If \(C'(t)=S(t)\) and \(S(t)>0\) throughout an interval, can \(C\) have an interior local maximum there?
- Why does \(g(t)=t^3\) show that a zero derivative is not sufficient for a local extremum?
- Can a differentiable function have a strict local extremum even if its derivative takes both signs arbitrarily close to that point? Explain briefly.
