Three Kinds of Information, Three Different Questions
A graph-analysis prompt may give values of \(f\), signs or values of \(f'\), and signs or values of \(f''\). These facts describe related but distinct features of the graph. A value \(f(a)\) tells you the graph’s height at \(x=a\). A value \(f'(a)\) tells you its slope there. A sign of \(f''\) on an interval tells you how the slopes change across that interval.
A reliable response keeps those meanings separate before combining them. In particular, the sign of \(f'(a)\) is not the sign of \(f(a)\), and the sign of \(f''\) does not by itself determine whether the function is increasing or decreasing.
These sign conditions concern an entire interval. A derivative value at one input, such as \(f'(a)=2\), describes the slope at that one point; by itself it does not establish that the function increases on a surrounding interval. Likewise, \(f''(a)>0\) describes the second derivative at \(a\), but does not establish concavity throughout an interval unless the sign is known across that interval.
What Each Type of Data Tells You
| Information supplied | Conclusion it supports | What it does not establish by itself |
|---|---|---|
| \(f(a)=k\) | The graph passes through \((a,k)\). | The slope or concavity at \(a\). |
| \(f'(a)=m\) | The tangent slope at \(x=a\) is \(m\). | The function’s height or its behavior on a whole interval. |
| \(f'(x)>0\) on an interval | The function increases on that interval. | Whether its values are positive or negative. |
| \(f''(x)>0\) on an interval | The graph is concave up; its slopes are increasing. | Whether the function itself increases or is positive. |
When values are given at two inputs, monotonicity also gives a consistency check. If \(f\) is increasing on an interval containing \(a<b\), then \(f(a)<f(b)\); if it is decreasing, then \(f(a)>f(b)\). A table that contradicts this relationship may have been misread, or the asserted derivative sign may not hold on the whole interval.
Why Derivative Signs Give These Conclusions
The connection between derivative signs and monotonicity follows from the Mean Value Theorem. If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), the theorem guarantees a number \(c\) with \(a<c<b\) such that the change in the function equals the derivative at \(c\) times the interval length.
If \(f'(x)>0\) throughout \((a,b)\), then \(f'(c)>0\) and \(b-a>0\). Therefore \(f(b)-f(a)>0\), so \(f(b)>f(a)\). The same reasoning applies to any pair of points in the interval, proving that \(f\) is increasing there. If \(f'(x)<0\) throughout, the difference is negative, so the function is decreasing.
Concavity follows by applying the Mean Value Theorem to \(f'\). If \(f''>0\) throughout an interval, then \(f'\) is increasing there: for any \(a<b\), the theorem applied to \(f'\) gives a point \(d\) between them such that \(f'(b)-f'(a)=f''(d)(b-a)>0\). Thus later tangent slopes are greater than earlier ones, which is the slope-based meaning of concave up. When \(f''<0\), the slopes decrease, so the graph is concave down.
For a twice-differentiable function, this also agrees with the chord description of concavity. If \(a<x<b\) and \(f'\) is increasing, the Mean Value Theorem gives a point \(u\) between \(a\) and \(x\) and a point \(v\) between \(x\) and \(b\) for which
Because \(u<v\), increasing \(f'\) gives \(f'(u)\le f'(v)\). Rearranging this inequality shows that \(f(x)\) is no greater than the value at \(x\) on the straight chord from \((a,f(a))\) to \((b,f(b))\). Thus the graph lies below its chords, as expected for a concave-up function.
Worked Example: Interpreting Values and Both Derivatives
Worked Example: Interpreting Values and Both Derivatives
Let \(f(x)=x^3\) for \(-1\le x\le 1\). Find the function values and derivative signs at the listed inputs, then describe the graph’s behavior on either side of \(0\).
First calculate the derivatives:
Evaluate the function and derivatives at the three inputs:
For example, at \(x=-1\), \(f(-1)=(-1)^3=-1\), \(f'(-1)=3(-1)^2=3\), and \(f''(-1)=6(-1)=-6\). At \(x=0\), all three quantities are \(0\). At \(x=1\), they are \(1\), \(3\), and \(6\), respectively.
On both \((-1,0)\) and \((0,1)\), \(f'(x)=3x^2>0\), so \(f\) is increasing on each interval. On \((-1,0)\), \(f''(x)=6x<0\), so the graph is concave down. On \((0,1)\), \(f''(x)=6x>0\), so it is concave up. The point values confirm the height information: the graph passes through \((-1,-1)\), \((0,0)\), and \((1,1)\).
At \(x=-1\), the function value is negative while the derivative is positive. This is not a contradiction: the graph is below the horizontal axis there but has a positive tangent slope. Differentiating \(x^3\) again verifies the calculations, so the signs and values are consistent.
Worked Example: Combining an Interval Sign Chart with Values
Worked Example: Combining an Interval Sign Chart with Values
Consider \(f(x)=\frac{x^3}{3}-x\) on \([-2,2]\). Use derivative signs to describe where the function increases, decreases, and is concave up or down. Then check the supplied point values.
Differentiate:
The sign of \(f'\) is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). The sign of \(f''\) is negative for \(x<0\) and positive for \(x>0\). Therefore \(f\) increases on \((-2,-1)\) and \((1,2)\), decreases on \((-1,1)\), is concave down on \((-2,0)\), and is concave up on \((0,2)\).
Now calculate the stated values:
These values agree with the derivative signs. From \(-2\) to \(-1\), the values rise from \(-\frac{2}{3}\) to \(\frac{2}{3}\); from \(-1\) to \(1\), the values fall from \(\frac{2}{3}\) to \(-\frac{2}{3}\); and from \(1\) to \(2\), they rise from \(-\frac{2}{3}\) to \(\frac{2}{3}\). The concavity statements concern the change in slope, not whether each value is above or below zero. Direct differentiation confirms both derivative formulas.
Worked Example: A Tangent Slope Is Not an Interval Trend
Worked Example: A Tangent Slope Is Not an Interval Trend
Let \(f(x)=\frac14(x-1)^2-2(x-1)+4\) on \([1,3]\). Find the height and slope at \(x=1\), and describe the function on \((1,3)\).
At \(x=1\), the terms containing \(x-1\) are zero, so \(f(1)=4\). Differentiate to obtain
Thus \(f'(1)=\frac12(0)-2=-2\), while \(f''(x)=\frac12>0\) throughout \((1,3)\). The graph has height \(4\) and tangent slope \(-2\) at \(x=1\); it is concave up throughout the interval. To determine whether it increases or decreases across the interval, examine \(f'\) there. Since \(1<x<3\), we have \(0<x-1<2\), so \(0<\frac12(x-1)<1\). Therefore \(-2<f'(x)<-1\), and \(f'(x)\) remains negative. The function decreases on \((1,3)\).
As a numerical check, \(f(3)=\frac14(2)^2-2(2)+4=1-4+4=1\), which is less than \(f(1)=4\), consistent with the decreasing behavior. Differentiating the original expression gives the displayed \(f'\) and \(f''\), confirming the slope and concavity calculations.
A Common Misreading
A common error is to treat positive second derivative as evidence that the function is increasing. For example, take \(f(x)=x^2\) on \((-2,-1)\). Then
The function is decreasing on this interval because \(f'<0\), even though it is concave up because \(f''>0\). Its endpoint values show the decrease: \(f(-2)=4\) and \(f(-1)=1\). The second derivative describes how slopes change, not whether those slopes are positive. Here the slopes are negative but increasing: for instance, \(f'(-2)=-4\) and \(f'(-1)=-2\).
Another mistake is to infer a function’s sign from its derivative. In the first example, \(f(-1)=-1\) but \(f'(-1)=3\). A positive slope does not mean the graph is above the horizontal axis. These errors are avoided by translating each datum separately before drawing conclusions.
A Reliable Response Process
Translate \(f(a)=k\) into the point \((a,k)\) on the graph. Do not infer a slope from the height.
A derivative value gives a tangent slope at one input. A positive or negative sign throughout an interval gives increasing or decreasing behavior there.
A positive or negative sign throughout an interval gives concave up or concave down behavior, respectively.
Compare interval behavior with any given endpoint or table values. State only conclusions supported by the locations and domains of the data.
Precise graph-analysis writing names the evidence and the conclusion together. For example: “Because \(f'(x)<0\) for \(1<x<3\), \(f\) is decreasing on \((1,3)\). Because \(f''(x)>0\) there, its graph is concave up.” This makes clear why the claims follow and keeps height, slope, and concavity distinct.
Check Your Understanding
For each question, identify exactly which function or derivative information supports your conclusion.
- If \(f(2)=-3\) and \(f'(2)=4\), what point is on the graph, and what is the tangent slope there?
- If \(f'(x)<0\) throughout \((0,5)\), what can you conclude about the function on that interval?
- If \(f''(x)>0\) throughout an interval but \(f'(x)<0\) there as well, describe both the direction of change and the concavity.
- For \(f(x)=x^2\) on \((-2,-1)\), calculate the signs of \(f'\) and \(f''\), then state the graph’s behavior.
- A table gives \(f(a)=2\), \(f(b)=5\), and says \(f'<0\) throughout \((a,b)\), where \(a<b\). Explain why these data cannot all be correct under the stated differentiability conditions.
