Connecting f, f' and f'' — AP® Calculus AB Notes

This topic pulls Unit 5 together. You are given one of $f$ , $f'$ or $f''$ , as a graph, a table or a formula, and asked about the others: where $f$ rises, where it peaks, how it bends, whether a tangent line over- or underestimates. Every year one free-response question is built this way around a graph of $f'$ . The skill is graphical analysis: reading the right feature of the right function and saying so by name.

One story told three ways

Three stacked graphs on the same x-axis from 0 to 2 pi: f(x) = sin x, f'(x) = cos x and f''(x) = -sin x. Dashed lines at pi/2, pi, 3pi/2 and 2pi line up the features: f peaks at pi/2 where f' is zero and changes from positive to negative; f has an inflection point at pi where f' is lowest and f'' changes sign; f'' is negative on (0, pi), where f is concave down. — **Figure 1.** $f(x) = \sin x$ , $f'(x) = \cos x$ and $f''(x) = -\sin x$ , lined up. Each feature of $f$ shows up as a zero or a turn one level down. LevlPrep original diagram.

Behavior of $f$	Read from $f'$	Read from $f''$
increasing	$f' > 0$	(not enough)
relative maximum	$f'$ changes from $+$ to $-$	$f'(c) = 0$ and $f''(c) < 0$
concave up	$f'$ increasing	$f'' > 0$
point of inflection	$f'$ changes from increasing to decreasing, or back	$f''$ changes sign

Always ask which function the given information is about. "Positive" means something different for $f$ , $f'$ and $f''$ .

Working from a graph of $f'$

Worked example 1. The graph of $f'(x) = x^3 - 3x^2 + 2$ on $[-1, 3.2]$ has a high point at $x = 0$ , a low point at $x = 2$ , and crosses the axis at $x \approx -0.73$ , $x = 1$ and $x \approx 2.73$ . Also $f(0) = 4$ .

Relative maximum of $f$ : at $x = 1$ , because $f'$ changes from positive to negative there.

Concave up: on $(-1, 0)$ and $(2, 3.2)$ , because $f'$ is increasing there.

Inflection points: at $x = 0$ and $x = 2$ , because $f'$ changes from increasing to decreasing at 0 and from decreasing to increasing at 2.

At $x = 2$ : $f'(2) = -2 < 0$ , so $f$ is decreasing there, and it has a point of inflection.

Tangent line at $x = 0$ : point $(0, 4)$ , slope $f'(0) = 2$ , so $y = 4 + 2x$ .

Notice the pattern in every justification: it names $f'$ and says what $f'$ does. "The graph changes direction at 1" names nothing and earns nothing.

Working from a table

A table gives exact values at a few points. Use the tests that need only those values, and the theorems that connect them.

Worked example 2. $f$ is twice differentiable and $f''$ is continuous.

$x$	$1$	$3$	$4$
$f(x)$	$1$	$4$	$2$
$f'(x)$	$0$	$0$	$-4$
$f''(x)$	$2$	$-3$	$-1$

At $x = 1$ : $f'(1) = 0$ and $f''(1) > 0$ , so by the Second Derivative Test $f$ has a relative minimum.

Inflection in $(1, 3)$ : $f''$ is continuous, $f''(1) > 0$ and $f''(3) < 0$ , so $f''$ changes from positive to negative somewhere in $(1, 3)$ : the graph has a point of inflection there.

Approximating $f(4.2)$ : $y = 2 - 4(x - 4)$ gives $f(4.2) \approx 1.2$ . Because $f''(4) < 0$ and $f''$ is continuous, $f$ is concave down near 4, the graph lies below the tangent line, and $1.2$ is an overestimate.

Working from words

Contexts describe derivatives in words. Translate before you calculate:

"Increasing at a decreasing rate": $f' > 0$ and $f'' < 0$ .
"Decreasing, but more and more slowly": $f' < 0$ and $f'' > 0$ .
"The rate of change is greatest": $f'$ has a maximum, which is where $f''$ changes from positive to negative, an inflection point of $f$ .
For motion, position $s$ : velocity $s'$ and acceleration $s''$ . With $s' > 0$ and $s'' < 0$ , the object moves forward and slows down.

A global fact from $f''$

Sometimes the sign of $f''$ holds everywhere, and that gives a global answer. If $f''(x) < 0$ for all $x$ and $f'(2) = 0$ , then $f'$ is decreasing on the whole line: positive before 2 and negative after. So $f$ rises to $x = 2$ and falls after, and $f(2)$ is the absolute maximum, not just a relative one.

The two classic swaps

Zeros versus turns. On a graph of $f'$ , zeros (where it crosses the axis) are candidates for extrema of $f$ ; turns (its peaks and valleys) are inflection points of $f$ .
Sign versus direction. The sign of $f'$ gives the direction of $f$ ; the direction of $f'$ gives the concavity of $f$ . "f is increasing because $f'$ is increasing" mixes them up.