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Monotonicity and Local Extrema · Tutorial 345 of 999

Meaning of Derivative Signs on Intervals

Interpret the derivative as a local rate of change, then distinguish a sign at one point from a sign condition across an interval.

Intermediate 9 min read

What You'll Learn

  • Interpret positive and negative derivatives as local tangent directions and rates of change.
  • Explain precisely what a nonzero derivative implies about nearby function values relative to the point.
  • Understand why a zero derivative does not by itself identify a local maximum or minimum.
  • Distinguish a derivative sign at one point from monotonic behavior across an interval.
  • Recognize the hypotheses needed to use an interval-wide derivative sign to conclude monotonicity.

What a Derivative Sign Tells You

A derivative describes how a function changes at a particular input. Its sign gives the direction of the function’s tangent line at that point: a positive derivative corresponds to an upward-sloping tangent, a negative derivative to a downward-sloping tangent, and a zero derivative to a horizontal tangent.

These are useful local interpretations, but they must be used carefully. A derivative sign at one point does not, by itself, tell you whether the function increases or decreases throughout a whole interval. To make that broader conclusion, information across the interval is needed.

Definition: If \(f\) is defined on an open interval containing \(c\), its derivative at \(c\), when it exists, is
$$ f'(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}. $$
The derivative is the instantaneous rate of change at \(c\), and geometrically it is the slope of the tangent line to the graph at \((c,f(c))\).

The quotient in the definition is a secant slope: it compares the function values at \(c\) and \(c+h\). Taking the limit as \(h\) approaches zero gives the tangent slope. If the derivative has units, those units are output units divided by input units—for example, meters per second for position measured in meters as a function of time in seconds.

Positive and Negative Derivatives

Suppose \(f'(c)>0\). The tangent line at \(c\) rises from left to right. More precisely, for inputs sufficiently close to \(c\), values to the right of \(c\) are above \(f(c)\), while values to the left are below \(f(c)\). This compares nearby values with the value at \(c\); it does not yet compare every pair of nearby inputs.

If \(f'(c)<0\), the tangent line falls from left to right. Near \(c\), values to the right are below \(f(c)\), while values to the left are above it. In either case, the derivative’s sign describes the function’s local direction as it passes through the point.

Key interpretation: A positive derivative at \(c\) means the function is moving upward through its value at \(c\), in the local tangent sense. A negative derivative means it is moving downward through that value. Neither statement alone guarantees that the function is increasing or decreasing across a specified interval.

Why the Sign Gives a Local Comparison

The local comparison follows directly from the derivative definition. When \(f'(c)>0\), the difference quotient is positive for all sufficiently small nonzero \(h\). Since \(h\) is positive on the right of \(c\) and negative on the left, the change \(f(c+h)-f(c)\) has the same sign as \(h\). A negative derivative reverses those comparisons.

1
Start with a positive derivative.
Let \(L=f'(c)>0\). Because the difference quotient approaches \(L\), it is greater than \(L/2>0\) whenever \(h\) is sufficiently close to zero and nonzero.
2
Use the sign of \(h\).
If \(h>0\), a positive quotient means \(f(c+h)-f(c)>0\). If \(h<0\), a positive quotient means \(f(c+h)-f(c)<0\).
3
Read the conclusion relative to \(f(c)\).
Nearby values to the right are above \(f(c)\); nearby values to the left are below \(f(c)\). This conclusion concerns comparisons with \(f(c)\), not every pair of inputs in a neighborhood.

For \(f'(c)<0\), the quotient is negative sufficiently close to \(c\). Multiplying its sign by the sign of \(h\) gives the reversed conclusions: values just to the right are below \(f(c)\), and values just to the left are above it.

Worked Example: Finding a Tangent’s Direction

Worked Example: Finding a Tangent’s Direction

Let \(f(x)=2x^2-x+3\). Find and interpret the derivative at \(x=2\).

Differentiate:

$$f'(x)=4x-1.$$
Substitute \(x=2\):
$$f'(2)=4(2)-1=8-1=7.$$
The derivative is positive, so the tangent line rises from left to right at \(x=2\). The function value there is
$$f(2)=2(2^2)-2+3=8-2+3=9.$$
The tangent line through \((2,9)\) with slope \(7\) is
$$y-9=7(x-2),\qquad y=7x-5.$$
Check the slope from the derivative definition:
$$\frac{f(2+h)-f(2)}{h}=\frac{9+7h+2h^2-9}{h}=7+2h\quad (h\ne 0),$$
which approaches \(7\) as \(h\to0\). The tangent slope agrees with \(f'(2)=7\).

A Zero Derivative Is Not a Verdict

If \(f'(c)=0\), the tangent line is horizontal. The derivative definition says that the average rate of change over a very short interval around \(c\) approaches zero. Equivalently, the function’s change is smaller than a nonzero linear change in \(h\) as \(h\) approaches zero. But a horizontal tangent alone does not say what the graph does on either side.

A zero derivative can occur at a local maximum, a local minimum, or a point where the function keeps increasing or decreasing. It may also occur at a point with more complicated nearby behavior. The derivative value by itself does not determine which case applies.

Worked Example: Two Different Behaviors with Zero Derivative

Compare \(f(x)=x^3\) and \(g(x)=x^2\) at \(x=0\).

For \(f\), use the derivative definition:

$$f'(0)=\lim_{h\to0}\frac{h^3-0}{h}=\lim_{h\to0}h^2=0.$$
Yet \(f(x)=x^3\) is negative for \(x<0\) and positive for \(x>0\). It passes through \((0,0)\) while increasing; it has neither a local maximum nor a local minimum at zero.

For \(g\):

$$g'(0)=\lim_{h\to0}\frac{h^2-0}{h}=\lim_{h\to0}h=0.$$
But \(g(x)=x^2\ge 0=g(0)\) for every \(x\), so \(g\) has a local minimum at zero.

Check the derivative values by differentiating: \(f'(x)=3x^2\), so \(f'(0)=0\), and \(g'(x)=2x\), so \(g'(0)=0\). The same derivative value at the point occurs with different local behavior.

A Pointwise Sign Is Not an Interval-Wide Sign

A function is increasing on an interval if \(x_1<x_2\) in that interval implies \(f(x_1)<f(x_2)\). It is decreasing if \(x_1<x_2\) implies \(f(x_1)>f(x_2)\). These definitions compare all ordered pairs of inputs in the interval.

By contrast, \(f'(c)>0\) is information at one input \(c\). It does not tell you what \(f'(x)\) is at other inputs. Even if it gives a local comparison with \(f(c)\), it does not establish all the pairwise comparisons required for increase on a larger interval.

Worked Example: A Positive Derivative at One Point Does Not Prove Interval Increase

Consider \(f(x)=x^3-3x\) and the interval \([-2,3]\). At \(x=2\),

$$f'(x)=3x^2-3,\qquad f'(2)=3(2^2)-3=12-3=9>0.$$
So the tangent at \(x=2\) rises from left to right. But the function is not increasing on the whole interval: \(-1<0\), while
$$f(-1)=(-1)^3-3(-1)=-1+3=2,$$
and
$$f(0)=0^3-3(0)=0.$$
Thus \(f(-1)=2>0=f(0)\), contrary to what increase on \([-2,3]\) would require. The positive derivative at \(2\) correctly describes the tangent there, but it cannot settle the behavior across the entire interval.

Check: differentiating \(x^3-3x\) gives \(3x^2-3\), and substituting \(2\) gives \(9\) as stated. The two function values also directly verify the failure of interval increase.

What an Interval-Wide Sign Can Establish

To conclude that a function increases or decreases across an interval, the derivative’s behavior throughout that interval matters. The Mean Value Theorem makes this distinction precise: if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for any \(x_1<x_2\) in \([a,b]\), it relates the secant slope between those inputs to a derivative at some point between them. Consequently, a positive derivative at every interior point guarantees increase on the interval; a negative derivative at every interior point guarantees decrease.

The requirements cover an interval, not just one input. A derivative sign at one point cannot replace the condition across all interior points. Also, a nonnegative derivative throughout an interval gives nondecreasing behavior, which allows equal function values; it does not necessarily give strict increase.

Keep the scopes separate: \(f'(c)>0\) is a pointwise statement about the tangent and nearby comparisons with \(f(c)\). A conclusion that \(f\) increases on an interval requires information across the interval together with the appropriate continuity and differentiability conditions.

Common Mistake: Extending One Tangent’s Direction Too Far

A common error is to calculate \(f'(c)>0\) and immediately write, “Therefore, \(f\) is increasing on \([a,b]\).” That conclusion does not follow from a single derivative value. For the previous example, the incorrect argument would be:

$$ f'(2)=9>0 \quad\Longrightarrow\quad f\text{ is increasing on }[-2,3]. $$

The implication is false: \(-1<0\), but \(f(-1)=2>0=f(0)\). The correction is to state only what the calculation supports: the tangent at \(x=2\) has positive slope. To claim interval increase, check the derivative sign throughout the relevant interval and verify the theorem’s hypotheses.

Why These Interpretations Matter

Derivative signs are central to analyzing graphs and rates. In an applied setting, a positive rate of change at a particular time means the measured quantity is increasing at that instant; it does not guarantee that it kept increasing over an entire time period. A zero rate identifies an instant with horizontal tangent, but it does not by itself identify a turning point.

Keeping the pointwise and interval-wide claims distinct prevents unsupported conclusions. First interpret the derivative where it is evaluated. Then, if the question asks about an interval, use information that applies across that interval.

Check Your Understanding

Answer each question by distinguishing what is known at one point from what would be needed for an interval-wide conclusion.

  1. If \(f'(c)<0\), what does this say about the tangent at \(c\), and how do sufficiently nearby values compare with \(f(c)\) on the left and right?
  2. Give a reason that \(f'(c)=0\) does not necessarily mean \(f\) has a local maximum or minimum at \(c\).
  3. For \(f(x)=x^3-3x\), calculate \(f'(2)\). Does that one value prove that \(f\) increases on \([-2,3]\)? Explain.
  4. What kind of derivative information is needed to use the Mean Value Theorem to conclude that a function increases across an interval?
  5. How does nondecreasing behavior differ from strictly increasing behavior?

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