4.5 Derivatives and the Shape of a Graph

Big idea

Critical points give you a short list of places where something might happen. This section supplies the screening. Two pieces of information do the work: the sign of the first derivative, which says whether the graph is rising or falling, and the sign of the second derivative, which says which way it bends.

The first of these follows from the mean value theorem. If $f' > 0$ across an interval then any two points in it satisfy $f(y) - f(x) = f'(c)(y - x) > 0$ when $y > x$, so $f$ is increasing there. So a sign chart for $f'$ is a complete description of where $f$ climbs and where it falls, and a critical point where the sign flips from positive to negative is a local maximum, no further argument needed.

The second derivative adds the curvature. Saying $f'' > 0$ says the slope itself is increasing, so the tangent lines keep tilting counterclockwise and the graph bends upward, holding water. Where the bending switches direction you get an inflection point, which is a feature of the graph in its own right, invisible to the first derivative. A population curve’s inflection point is the moment growth stops accelerating and starts easing off, which is often the number people actually want.

The two tests overlap but are not equivalent. The second derivative test is quick when it works and silent when $f''$ happens to be zero at the critical point. The first derivative test always works, at the cost of a sign chart. Learning when to reach for each is most of the practical content here.

Decoder

At a critical point, a sign change of $f'$ from positive to negative marks a local maximum and a change from negative to positive marks a local minimum; if $f'$ keeps its sign, the point is neither.

The phrase that carries the content is “sign change.” The value of $f'$ at the critical point itself is not what decides; it is the behavior on the two sides. That is why the test survives at points where $f'$ does not exist, such as a corner or a cusp, where the second derivative test has nothing to say.

To run it, mark the critical points and any points where $f$ is undefined on a number line, then test one convenient input in each resulting interval. Only the sign of $f'$ matters, so factor the derivative and read the signs off the factors rather than computing values.

The classic mistake is treating the second derivative test as decisive when it returns zero. If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive, and that is a genuine gap, not a hint. At $x = 0$ the functions $x^4$, $-x^4$ and $x^3$ all have first and second derivatives equal to zero, and they have a minimum, a maximum, and no extremum respectively. Fall back to the first derivative test.

Definitions and results

Increasing and decreasing. $f$ is increasing on an interval if $x_1 < x_2$ implies $f(x_1) < f(x_2)$ there, and decreasing if the inequality reverses.

The monotonicity test. If $f' > 0$ on an interval then $f$ is increasing on it; if $f' < 0$ then $f$ is decreasing. This is a corollary of the mean value theorem.

The first derivative test. Let $c$ be a critical point of a continuous $f$. If $f'$ changes from positive to negative at $c$, then $f(c)$ is a local maximum. If $f'$ changes from negative to positive, $f(c)$ is a local minimum. If $f'$ has the same sign on both sides, $f(c)$ is neither.

Concavity. $f$ is concave up on an interval where $f'$ is increasing, and concave down where $f'$ is decreasing. Equivalently, a concave up graph lies above each of its tangent lines on that interval, and a concave down graph lies below them.

The concavity test. If $f''(x) > 0$ on an interval then $f$ is concave up there; if $f''(x) < 0$ then $f$ is concave down.

Inflection point. A point on the graph where the concavity changes. Since concavity is governed by the sign of $f''$, look where $f''(x) = 0$ or $f''$ fails to exist, then confirm that the sign actually flips. A zero of $f''$ with no sign change is not an inflection point, as $f(x) = x^4$ shows at the origin.

The second derivative test. Suppose $f'(c) = 0$ and $f''$ is continuous near $c$. If $f''(c) > 0$ then $f$ has a local minimum at $c$; if $f''(c) < 0$ then $f$ has a local maximum at $c$. If $f''(c) = 0$ the test gives no information.

Reading the two signs together. The four combinations of signs describe four shapes: rising and concave up, rising and concave down, falling and concave up, falling and concave down. Knowing both at a point pins down the local picture completely.

Worked examples

A full analysis of a cubic

Analyze $f(x) = 2x^3 - 3x^2 - 12x + 1$.

Differentiate twice:

$$ f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1), \qquad f''(x) = 12x - 6 $$

The critical points are $x = -1$ and $x = 2$. For the sign chart, test $x = -2$, $x = 0$, $x = 3$ in the factored first derivative: the factors give positive, negative, positive. So $f$ increases on $(-\infty,-1)$, decreases on $(-1,2)$, and increases on $(2,\infty)$.

By the first derivative test, $x = -1$ is a local maximum with $f(-1) = -2 - 3 + 12 + 1 = 8$, and $x = 2$ is a local minimum with $f(2) = 16 - 12 - 24 + 1 = -19$.

The second derivative test agrees and is faster: $f''(-1) = -18 < 0$ gives the maximum, and $f''(2) = 18 > 0$ gives the minimum.

For concavity, $f''(x) = 0$ at $x = \tfrac12$, with $f'' < 0$ to the left and $f'' > 0$ to the right. So the graph is concave down on $(-\infty,\tfrac12)$, concave up on $(\tfrac12,\infty)$, and has an inflection point at $\big(\tfrac12, -\tfrac{11}{2}\big)$. Notice the inflection sits midway between the two turning points, as it always does for a cubic.

Concavity and inflection in a quartic

Analyze the shape of $f(x) = x^4 - 6x^2$.

Differentiate:

$$ f'(x) = 4x^3 - 12x = 4x(x^2 - 3), \qquad f''(x) = 12x^2 - 12 = 12(x-1)(x+1) $$

The critical points are $x = 0$ and $x = \pm\sqrt{3}$. The second derivative test settles all three: $f''(0) = -12 < 0$, so $x = 0$ is a local maximum with value $0$; $f''(\pm\sqrt{3}) = 36 - 12 = 24 > 0$, so both are local minima with value $9 - 18 = -9$.

For concavity, $f'' > 0$ when $|x| > 1$ and $f'' < 0$ when $|x| < 1$. The sign changes at both $x = -1$ and $x = 1$, so both are inflection points, at height $1 - 6 = -5$. The graph is a symmetric W: up, down through an inflection, over the hump at the origin, down through the second inflection, and up again.

A critical point with no second derivative

Analyze $f(x) = x^{2/3}$ at the origin.

Here $f'(x) = \frac{2}{3}x^{-1/3}$, which is undefined at $x = 0$, so $x = 0$ is a critical point. The second derivative test cannot be used, since $f''(x) = -\frac{2}{9}x^{-4/3}$ does not exist there either.

The first derivative test handles it without trouble. For $x < 0$ the cube root is negative, so $f' < 0$; for $x > 0$ it is positive, so $f' > 0$. The sign changes from negative to positive, so $f$ has a local minimum at $x = 0$ with value $0$. Check directly: $x^{2/3} = \big(x^{1/3}\big)^2 \ge 0$ for every $x$, with equality only at $0$, so this is even an absolute minimum.

Meanwhile $f'' < 0$ on both sides of the origin, so the graph is concave down on each side and there is no inflection point, despite the sharp change in direction. Concavity and direction are independent pieces of information.

The inconclusive case

Compare $f(x) = x^4$ and $g(x) = x^3$ at the origin.

Both satisfy $f'(0) = 0$ and $f''(0) = 0$, so the second derivative test says nothing about either. The first derivative test separates them. For $f$, the derivative $4x^3$ is negative on the left and positive on the right, so there is a local minimum at $0$. For $g$, the derivative $3x^2$ is positive on both sides, so $g$ is increasing through the origin and there is no extremum.

The moral is procedural: a zero second derivative at a critical point is a signal to switch tests, not a verdict.

Practice

Begin with sign charts. These drills ask where a function increases and decreases, and then use the direction of the sign change to classify each critical point.

Next the bending. Here you find where $f''$ is positive and negative, name the intervals of concavity, and confirm which candidates are genuine inflection points.

Finally the quick test. These give you a critical point and ask you to classify it from the sign of the second derivative, including cases where the test comes back inconclusive.

Quiz

Six items across monotonicity, the first derivative test, concavity and inflection points, and the second derivative test.