Reading the graph of f′

A common AP Calculus question gives the graph of \(f'\) and asks about \(f\). The difficulty is mostly translation. The picture is not the graph of \(f\). It is information about \(f\). Two facts organize nearly everything that follows:

• the sign of \(f'\) tells you whether \(f\) is increasing or decreasing
• the slope of \(f'\) tells you the concavity of \(f\)

The translation table

You see on the graph of \(f'\)	You conclude about \(f\)
\(f'\) above the \(x\)-axis	\(f\) is increasing
\(f'\) below the \(x\)-axis	\(f\) is decreasing
\(f'\) crosses zero from positive to negative	\(f\) has a relative maximum
\(f'\) crosses zero from negative to positive	\(f\) has a relative minimum
\(f'\) is decreasing	\(f\) is concave down
\(f'\) is increasing	\(f\) is concave up
\(f'\) changes from decreasing to increasing, or the reverse	\(f\) may have a point of inflection
signed area between \(f'\) and the axis	total change in \(f\)

The height of the graph and the shape of the graph answer different questions. The height of \(f'\) tells you the sign of the derivative, and the shape of \(f'\) tells you whether the derivative itself is increasing or decreasing.

A worked example

Suppose the graph of \(f'\) on \([0,8]\) consists of line segments through \((0,3),\; (2,0),\; (4,-3),\; (6,0),\; (8,3)\).

Where is \(f\) increasing? Wherever \(f'>0\), so \(f\) increases on \((0,2)\) and \((6,8)\). It does not matter that \(f'\) is decreasing on \((0,2)\). The derivative is still positive there, so \(f\) is increasing.

Where does \(f\) have a relative maximum? At \(x=2\), because \(f'\) changes from positive to negative there. At \(x=6\), \(f'\) changes from negative to positive, so \(f\) has a relative minimum.

Where is \(f\) concave down? Where \(f'\) is decreasing, and that occurs on \((0,4)\). Likewise, \(f\) is concave up on \((4,8)\), where \(f'\) is increasing. The point \(x=4\) is a point of inflection because the concavity changes there.

Recovering values of \(f\)

If one value of \(f\) is known, signed area under \(f'\) gives the rest. Suppose \(f(0)=1\), so that

\[f(8) = f(0)+\int_0^8 f'(x)\,dx\]

From \(0\) to \(2\), the graph contributes a triangle of area \(3\). From \(2\) to \(6\), the graph is below the axis and contributes \(-6\). From \(6\) to \(8\), it contributes another \(3\), so \(f(8) = 1+3-6+3 = 1\). The function ends at the same value where it began, even though it increased, decreased, and increased again in between.

To find an absolute minimum, evaluate \(f\) at the relevant candidates. Here, \(f(0)=1,\; f(2)=4,\; f(6)=-2,\; f(8)=1\), so the absolute minimum occurs at \(x=6\). No formula for \(f\) was needed.

Three common mistakes

Treating the picture as \(f\)

The lowest point on the graph of \(f'\) is not automatically a minimum of \(f\). A minimum of \(f'\) says something about the slope of \(f\) and therefore about concavity, and extrema of \(f\) are found by looking for sign changes in \(f'\).

Assuming \(f'=0\) is enough

A zero of \(f'\) is only a critical point. For a relative maximum or minimum, the sign of \(f'\) must change, and if \(f'\) touches zero and remains positive on both sides, \(f\) keeps increasing.

Answering about the wrong function

It helps to label conclusions explicitly. Write “\(f\) is increasing because \(f'>0\)” rather than simply “increasing.”