How Derivatives Shape a Graph

Critical numbers mark where extrema can occur, but not which are maxima and which are minima. The two derivatives settle it. Since $f^{'} (x)$ is the slope of the curve, its sign records the direction of travel, and $f^{''} (x)$ records how that direction is turning. Reading these two signs off a function reconstructs its shape without plotting a single extra point.

Increasing, decreasing, and the sign of $f^{'}$

Between two points where the tangent slopes uphill the function is rising; between two points where it slopes downhill it is falling. The Mean Value Theorem promotes that observation to a theorem.

For any $x_{1} < x_{2}$ in the interval, the Mean Value Theorem gives a $c$ between them with $f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})$ . The factor $x_{2} - x_{1}$ is positive, so $f (x_{2}) - f (x_{1})$ takes the sign of $f^{'} (c)$ : positive slope makes $f (x_{2}) > f (x_{1})$ , negative slope reverses it.

The critical numbers cut the domain into intervals on which $f^{'}$ keeps a constant sign, so testing one point per interval settles the whole interval. For $f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 5$ ,

f^{'} (x) = 12 x^{3} - 12 x^{2} - 24 x = 12 x (x - 2) (x + 1),

with critical numbers $x = - 1, 0, 2$ . A sign chart multiplies the signs of the three factors on each interval.

Interval	$12 x$	$x - 2$	$x + 1$	$f^{'} (x)$	$f$
$x < - 1$	$-$	$-$	$-$	$-$	decreasing
$- 1 < x < 0$	$-$	$-$	$+$	$+$	increasing
$0 < x < 2$	$+$	$-$	$+$	$-$	decreasing
$x > 2$	$+$	$+$	$+$	$+$	increasing

A sign chart of

f^{'}

: each critical number is a boundary where the slope changes sign, and the arrows show the matching rise and fall of

f

The First Derivative Test

A local extremum sits where the function stops rising and starts falling, or the reverse. In terms of the derivative, an extremum is a sign change of $f^{'}$ .

The test follows from the Increasing/Decreasing Test: rising then falling makes a peak, falling then rising makes a valley, and no change of direction makes neither.

The four sign patterns of

f^{'}

around a critical number

c

, and the shape each produces.

For the function above, $f^{'}$ changes from negative to positive at $x = - 1$ and at $x = 2$ , so both give local minima, $f (- 1) = 0$ and $f (2) = - 27$ . It changes from positive to negative at $x = 0$ , so $f (0) = 5$ is a local maximum.

Concavity and the second derivative

The first derivative reports direction; the second reports how direction bends. A curve can rise while curving up like a bowl or up like a dome, and the two look completely different.

Concave upward means the slope $f^{'}$ is itself increasing, which happens exactly when $f^{''} > 0$ . Concave downward is the mirror image.

A point where concavity switches is where the curve changes the way it bends.

At an inflection point the second derivative changes sign, so any inflection point with $f^{''}$ defined has $f^{''} = 0$ there. The converse fails: $f (x) = x^{4}$ has $f^{''} (0) = 0$ but $(0, 0)$ is not an inflection point, since $f^{''} = 12 x^{2} \geq 0$ never changes sign.

Concave up (slopes increasing) meets concave down (slopes decreasing) at the inflection point, where the curve crosses its own tangent.

The Second Derivative Test

At a critical number where $f^{'} (c) = 0$ , the concavity settles which kind of extremum it is. A bowl-shaped graph sits above its horizontal tangent (minimum); a dome sits below it (maximum).

The test is inconclusive when $f^{''} (c) = 0$ : there might be a maximum, a minimum, or neither, and the First Derivative Test must be used instead. It is also inconclusive when $f^{''} (c)$ does not exist.

For $f (x) = x^{4} - 4 x^{3}$ , the derivatives are $f^{'} (x) = 4 x^{2} (x - 3)$ and $f^{''} (x) = 12 x (x - 2)$ . The critical numbers are $x = 0$ and $x = 3$ . At $x = 3$ , $f^{''} (3) = 36 > 0$ , so $f (3) = - 27$ is a local minimum. At $x = 0$ , $f^{''} (0) = 0$ and the Second Derivative Test gives nothing; the First Derivative Test shows $f^{'} < 0$ on both sides, so $x = 0$ yields neither. The second derivative vanishes at $x = 0$ and $x = 2$ , and it changes sign at both, so $(0, 0)$ and $(2, - 16)$ are inflection points.

Interval	$f^{''} (x) = 12 x (x - 2)$	concavity
$x < 0$	$+$	upward
$0 < x < 2$	$-$	downward
$x > 2$	$+$	upward

The two tests carry the same information in different form. The choice between them is one of convenience: the Second Derivative Test is a single evaluation when $f^{''}$ is easy, while the First Derivative Test always applies and handles the cases the second cannot.

The graph of

y = x^{4} - 2 x^{2}

: two local minima at

x = \pm 1

flank a local maximum at the origin, with inflection points where the concavity flips.

The correspondence between $f$ , $f^{'}$ , and $f^{''}$

Stacking a function above its first and second derivatives makes every feature line up vertically. Where $f^{'}$ crosses zero, $f$ has a horizontal tangent; where $f^{''}$ crosses zero, $f$ has an inflection.

f

f^{'}

, and

f^{''}

aligned: zeros of

f^{'}

mark extrema of

f

; zeros of

f^{''}

mark inflections of

f

and extrema of

f^{'}

Limits at infinity and horizontal asymptotes

The tests above describe local shape. End behavior asks a different question: what does $f (x)$ approach as $x$ runs off to $\pm \infty$ ?

A curve can approach its asymptote from above, from below, or by oscillating across it, and the two ends may settle toward different values. The base case is the reciprocal power.

x

grows,

1/ x^{r}

collapses toward the asymptote

y = 0

; the rational function settles toward its ratio-of-leading-coefficients line.

For a rational function, divide numerator and denominator by the highest power of $x$ in the denominator, then send each reciprocal power to zero. For $f (x) = \frac{3 x ^{2} - x - 2}{5 x ^{2} + 4 x + 1}$ ,

x \to \infty lim \frac{3 x ^{2} - x - 2}{5 x ^{2} + 4 x + 1} = x \to \infty lim \frac{3 - \frac{1}{x} - \frac{2}{x ^{2}}}{5 + \frac{4}{x} + \frac{1}{x ^{2}}} = \frac{3 - 0 - 0}{5 + 0 + 0} = \frac{3}{5} .

The same value comes out at $- \infty$ , so $y = \frac{3}{5}$ is a horizontal asymptote at both ends. The rule generalizes: for a rational function the horizontal asymptote is the ratio of leading coefficients when the degrees match, $y = 0$ when the denominator's degree is larger, and there is none when the numerator's degree is larger.

A radical in the numerator makes the two ends disagree, because $x^{2}$ equals $x$ for positive $x$ but $- x$ for negative $x$ .

Stewart, §3.3 — How Derivatives Affect the Shape of a Graph: the Increasing/Decreasing Test, First Derivative Test, Concavity Test, inflection points, and Second Derivative Test; §3.4 — Limits at Infinity; Horizontal Asymptotes: limits at infinity, horizontal asymptotes, vanishing reciprocal powers, and end behavior of rational functions. ↩

Increasing, decreasing, and the sign of f′

The First Derivative Test

Concavity and the second derivative

The Second Derivative Test

The correspondence between f, f′, and f′′

Limits at infinity and horizontal asymptotes

Footnotes

Increasing, decreasing, and the sign of $f^{'}$

The correspondence between $f$ , $f^{'}$ , and $f^{''}$