Reading the Three Graphs Together
A graph of a function shows its values. The graph of its first derivative shows the function’s slope at each input, and the graph of its second derivative shows how that slope changes. These are three different views of the same function, so their features cannot be chosen independently.
When matching proposed graphs, the aim is not merely to notice that each graph looks plausible on its own. The graphs must agree about where slopes are zero, where slopes are positive or negative, and where the function bends upward or downward. A zero or turning point in one graph can impose a specific condition on another.
What Each Graph Tells You
Suppose \(f\) is differentiable on an interval. If \(f'(x)>0\) throughout that interval, \(f\) is increasing there; if \(f'(x)<0\), \(f\) is decreasing there. A zero of \(f'\) means the graph of \(f\) has a horizontal tangent at that input. If \(f'\) changes from positive to negative, the function changes from increasing to decreasing and has a local maximum. A change from negative to positive gives a local minimum.
A zero of \(f'\) by itself does not guarantee a local maximum or minimum. The sign on either side matters. The function may continue increasing through a point where its tangent is horizontal.
If \(f\) is twice differentiable, \(f''(x)>0\) means its slopes are increasing, so the graph of \(f\) is concave up on that interval. If \(f''(x)<0\), its slopes are decreasing and the graph is concave down. A change in the sign of \(f''\) indicates a change in concavity; a zero of \(f''\) without a sign change does not establish one.
The same facts can be read one level up: \(f''(x)\) is the slope of the graph of \(f'\). Where \(f''>0\), the graph of \(f'\) is increasing; where \(f''<0\), it is decreasing. A sign change in \(f''\) therefore corresponds to a local extremum of \(f'\).
| Feature in a derivative graph | Constraint on the related graph |
|---|---|
| \(f'(x)>0\) | \(f\) is increasing at that input; on an interval of positivity, \(f\) is increasing throughout. |
| \(f'(x)<0\) | \(f\) is decreasing at that input; on an interval of negativity, \(f\) is decreasing throughout. |
| \(f'(a)=0\) | \(f\) has a horizontal tangent at \(a\), if it is differentiable there. |
| \(f''(x)>0\) or \(f''(x)<0\) | The graph of \(f\) is respectively concave up or concave down on an interval where the sign holds. |
| \(f''(a)=0\) | The graph of \(f'\) has a horizontal tangent at \(a\), if \(f'\) is differentiable there. A concavity change is not guaranteed. |
Why the Sign Tests Work
The sign tests follow from the Mean Value Theorem. If \(f\) is continuous on \([x,y]\) and differentiable on \((x,y)\), then there is a number \(c\) between \(x\) and \(y\) such that \(f(y)-f(x)=f'(c)(y-x)\). Since \(y-x>0\), the sign of the difference in function values is the sign of \(f'(c)\).
If \(f'(t)>0\) for every \(t\) in \((x,y)\), the Mean Value Theorem gives \(f(y)-f(x)=f'(c)(y-x)>0\). Thus \(f(y)>f(x)\), which is the definition of increasing behavior.
If \(f'(t)<0\) throughout \((x,y)\), then \(f(y)-f(x)=f'(c)(y-x)<0\). Thus \(f(y)<f(x)\), so \(f\) is decreasing.
When \(f'\) is continuous on \([x,y]\) and differentiable on \((x,y)\), the Mean Value Theorem applied to \(f'\) gives \(f'(y)-f'(x)=f''(c)(y-x)\). A positive \(f''\) makes \(f'\) increase; a negative \(f''\) makes it decrease.
For example, if \(f'\) is positive just to the left of \(a\) and negative just to the right, the first two sign conclusions show that \(f\) increases toward \(a\) and decreases after \(a\). Its value at \(a\) is therefore locally greater than nearby values: \(f\) has a local maximum. Reversing the signs gives a local minimum. These conclusions require the stated differentiability and sign behavior on the relevant intervals.
Worked Example: Matching a Cubic and Its Derivatives
Worked Example: Matching a Cubic and Its Derivatives
Take \(f(x)=x^3-3x\). Find and compare the zeros and signs of \(f\), \(f'\), and \(f''\), and use them to describe the graph relationships.
First find the zeros of \(f\) by factoring: \(f(x)=x(x^2-3)=x(x-\sqrt{3})(x+\sqrt{3})\). Thus the graph of \(f\) crosses the \(x\)-axis at \(x=-\sqrt{3},0,\sqrt{3}\).
Differentiate: \(f'(x)=3x^2-3=3(x-1)(x+1)\). Its zeros are \(x=-1\) and \(x=1\). The factorization shows that \(f'>0\) for \(x<-1\), \(f'<0\) for \(-1<x<1\), and \(f'>0\) for \(x>1\). Therefore \(f\) increases, then decreases, then increases. The sign changes give a local maximum at \(-1\) and a local minimum at \(1\).
Compute their heights: \(f(-1)=(-1)^3-3(-1)=-1+3=2\), and \(f(1)=1^3-3(1)=1-3=-2\). The corresponding turning points are \((-1,2)\) and \((1,-2)\).
Differentiate again: \(f''(x)=6x\). This is negative for \(x<0\) and positive for \(x>0\). So \(f\) is concave down to the left of zero and concave up to the right. The concavity changes at \(x=0\), where \(f(0)=0\), giving an inflection point at \((0,0)\).
The graph of \(f'\) is the parabola \(3x^2-3\): it crosses the horizontal axis at \(-1\) and \(1\), is below that axis between them, and is above it outside. The graph of \(f''\) is the line \(6x\), crossing zero at \(0\). Its negative-to-positive sign change means \(f'\) decreases before zero and increases after zero, so \(f'\) has a minimum at \(x=0\).
Check the derivative calculations: differentiating \(x^3-3x\) gives \(3x^2-3\), and differentiating \(3x^2-3\) gives \(6x\). Also, \(f'(-1)=3-3=0\), \(f'(1)=3-3=0\), and \(f''(0)=0\), as required by the identified horizontal tangents and slope change. The three graphs have consistent zeros, signs, turning behavior, and concavity.
Worked Example: A Quartic with Two Minima
Worked Example: A Quartic with Two Minima
Let \(g(x)=x^4-2x^2\). Determine how the zeros and signs of its derivatives constrain the shape of \(g\).
Differentiate: \(g'(x)=4x^3-4x=4x(x-1)(x+1)\). The derivative is zero at \(-1,0,1\). Testing the factors on the intervals gives \(g'<0\) for \(x<-1\), \(g'>0\) for \(-1<x<0\), \(g'<0\) for \(0<x<1\), and \(g'>0\) for \(x>1\). Thus \(g\) decreases, increases, decreases, and increases in succession.
The sign changes identify local minima at \(-1\) and \(1\), and a local maximum at \(0\). Their function values are \(g(-1)=1-2=-1\), \(g(0)=0\), and \(g(1)=1-2=-1\). So the points are \((-1,-1)\), \((0,0)\), and \((1,-1)\).
Now differentiate \(g'\): \(g''(x)=12x^2-4=4(3x^2-1)\). Its zeros satisfy \(3x^2-1=0\), so \(x^2=1/3\) and \(x=\pm 1/\sqrt{3}\). Since \(g''\) is positive outside these roots and negative between them, \(g\) is concave up for \(x<-1/\sqrt{3}\), concave down for \(-1/\sqrt{3}<x<1/\sqrt{3}\), and concave up for \(x>1/\sqrt{3}\).
At either candidate concavity-change input, \(x^2=1/3\) and \(x^4=1/9\). Therefore \(g(x)=1/9-2/3=1/9-6/9=-5/9\). The inflection points are \((-1/\sqrt{3},-5/9)\) and \((1/\sqrt{3},-5/9)\).
As a consistency check, \(g'\) decreases where \(g''<0\) and increases where \(g''>0\), so \(g'\) has a local maximum at \(-1/\sqrt{3}\) and a local minimum at \(1/\sqrt{3}\). Differentiating the original formula gives \(4x^3-4x\), and differentiating that expression gives \(12x^2-4\), confirming both derivative graphs used in the analysis.
Worked Example: A Horizontal Tangent Without an Extremum
Worked Example: A Horizontal Tangent Without an Extremum
Consider \(h(x)=(x-2)^3+1\). Determine what its derivative graphs say at \(x=2\), and decide whether that point is a maximum or minimum.
Differentiate: \(h'(x)=3(x-2)^2\). This derivative is zero at \(x=2\), but it is positive on both sides because a nonzero square is positive. Consequently, \(h\) is increasing on both sides of \(2\); there is no change from increasing to decreasing or from decreasing to increasing. The zero of \(h'\) gives a horizontal tangent, not a local extremum.
Differentiate once more: \(h''(x)=6(x-2)\). It is negative for \(x<2\) and positive for \(x>2\). Therefore the graph changes from concave down to concave up at \(x=2\). Since \(h(2)=(2-2)^3+1=1\), the point is \((2,1)\).
Check the derivatives: differentiating \((x-2)^3+1\) gives \(3(x-2)^2\), and differentiating that gives \(6(x-2)\). The first derivative is zero at \(2\) but has the same positive sign on both sides; the second derivative changes sign there. Hence the graph has a horizontal tangent and a concavity change, but not a maximum or minimum.
A Systematic Matching Method
When presented with three candidate graphs, begin with the proposed graph of \(f''\), then use it to check \(f'\), and finally use \(f'\) to check \(f\). A mismatch at either step rules out the proposed triple.
- Mark the zeros of \(f'\). Check that the proposed graph of \(f\) has horizontal tangents at those inputs.
- Record the sign of \(f'\) on each interval between its zeros. Check that \(f\) increases where that sign is positive and decreases where it is negative.
- At each zero of \(f'\), check the sign change. Positive to negative requires a local maximum; negative to positive requires a local minimum; no sign change means neither conclusion follows from this test.
- Mark the zeros and signs of \(f''\). Check that \(f'\) decreases where \(f''<0\) and increases where \(f''>0\).
- Where \(f''\) changes sign, check for a concavity change in \(f\). A zero without a sign change does not force one.
These checks concern relationships, not absolute heights. Adding a constant to \(f\) moves the graph of \(f\) vertically but leaves \(f'\) and \(f''\) unchanged. Therefore, derivative graphs determine slope and concavity behavior but do not by themselves determine the vertical position of \(f\).
Common Pitfall: Treating Every Zero as a Turning Point
A frequent error is to see \(f'(a)=0\) and immediately label \(a\) a maximum or minimum. That conclusion ignores the signs on either side. For instance, \(h'(x)=3(x-2)^2\) is zero at \(2\), but it is positive both before and after \(2\). The graph keeps increasing, so \(2\) is not a turning point.
Another error is to infer an inflection point solely from \(f''(a)=0\). The useful check is whether \(f''\) changes sign. A zero with no sign change does not show that concavity changes. In either case, compare the intervals on both sides rather than treating the zero itself as a complete conclusion.
Check Your Understanding
Use the relationships among the three graphs to answer these questions.
- If \(f'(x)<0\) on an interval, what must the graph of \(f\) do there?
- If \(f'\) changes from negative to positive at \(a\), what local behavior does \(f\) have at \(a\)?
- What does \(f'(a)=0\) guarantee, and what does it not guarantee?
- If \(f''\) is negative to the left of \(b\) and positive to the right, what happens to the concavity of \(f\) at \(b\)?
- Why does a zero of \(f''\) alone not establish that \(f\) has an inflection point?
