Applications of Derivatives - Key Properties

Concavity describes the curvature of a function's graph, providing insight into how the rate of change itself is changing. The second derivative test leverages this geometric information to classify critical points efficiently.

DefinitionConcavity

Let $f$ be differentiable on an interval $I$ .

$f$ is concave up on $I$ if $f'$ is increasing on $I$
$f$ is concave down on $I$ if $f'$ is decreasing on $I$

Geometrically, a function is concave up when its graph lies above its tangent lines, and concave down when it lies below them.

TheoremConcavity Test

Let $f$ be twice differentiable on an interval $I$ .

If $f''(x) > 0$ for all $x \in I$ , then $f$ is concave up on $I$
If $f''(x) < 0$ for all $x \in I$ , then $f$ is concave down on $I$

DefinitionInflection Points

A point $(c, f(c))$ is an inflection point of $f$ if the graph of $f$ changes concavity at $c$ . Typically, this occurs where $f''(c) = 0$ or $f''(c)$ does not exist, though not every such point is an inflection point.

ExampleFinding Inflection Points

Find the inflection points of $f(x) = x^4 - 4x^3$ .

Compute the derivatives: $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$ $f''(x) = 12x^2 - 24x = 12x(x - 2)$

Setting $f''(x) = 0$ : $x = 0$ or $x = 2$ .

Test concavity:

For $x < 0$ : $f''(−1) = 12(−1)(−3) = 36 > 0$ (concave up)
For $0 < x < 2$ : $f''(1) = 12(1)(−1) = −12 < 0$ (concave down)
For $x > 2$ : $f''(3) = 12(3)(1) = 36 > 0$ (concave up)

Concavity changes at both $x = 0$ and $x = 2$ , so both are inflection points.

TheoremSecond Derivative Test

Let $f$ be twice differentiable and suppose $f'(c) = 0$ .

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 is inconclusive

The second derivative test provides a quick way to classify critical points without analyzing sign changes in $f'$ .

ExampleUsing the Second Derivative Test

Classify the critical points of $g(x) = x^4 - 4x^3 + 10$ .

$g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$

Critical points: $x = 0$ and $x = 3$ .

$g''(x) = 12x^2 - 24x$

At $x = 0$ : $g''(0) = 0$ (inconclusive) At $x = 3$ : $g''(3) = 12(9) - 24(3) = 108 - 72 = 36 > 0$ (local minimum)

For $x = 0$ , we use the first derivative test. Since $g'$ doesn't change sign at $x = 0$ (it's negative before and after), there's no extremum at $x = 0$ .

Remark

Curve sketching combines information about:

Critical points (where $f' = 0$ or undefined)
Intervals of increase/decrease (sign of $f'$ )
Concavity (sign of $f''$ )
Inflection points (where $f''$ changes sign)
Asymptotes and limits at infinity

Together, these provide a complete picture of a function's behavior.

ExampleComplete Curve Sketch

Sketch $f(x) = \frac{x^2}{x^2 - 4}$ .

Domain: $x \neq \pm 2$ Vertical asymptotes: $x = \pm 2$ Horizontal asymptote: $\lim_{x \to \pm\infty} f(x) = 1$

$f'(x) = \frac{2x(x^2-4) - x^2(2x)}{(x^2-4)^2} = \frac{-8x}{(x^2-4)^2}$

Critical point: $x = 0$ (local maximum since $f'$ changes from positive to negative)

The second derivative analysis shows inflection points exist but are complex to compute. Combined with the asymptotic behavior, this information suffices for a qualitative sketch.