Applications of Derivatives and Extrema
The first derivative is a powerful tool for analysing the shape of a graph. This page follows Walz, Foundations of Mathematics, the first part of Section 6.2: it connects the sign of the derivative to whether a function rises or falls, and uses the first and second derivatives to locate the largest and smallest values of a function — its extrema.
Throughout we build on the derivatives and rules developed on the page Derivatives and Derivative Rules.
Monotonicity and the Sign of the First Derivative
A function is monotonically increasing if $x_1 < x_2$ always implies $f(x_1) \le f(x_2)$, and strictly monotonically increasing if $x_1 < x_2$ implies $f(x_1) < f(x_2)$; the decreasing cases are defined analogously. For differentiable functions there is a simple criterion in terms of the derivative.
Monotonicity and the sign of $f'$. A differentiable function $f$ is monotonically increasing if and only if $f'(x) \ge 0$ for all $x$, and monotonically decreasing if and only if $f'(x) \le 0$ for all $x$.
For example $g(x) = x^2 - 2x$ has $g'(x) = 2x - 2$, so $g$ is decreasing for $x \le 1$ (there $g'(x) \le 0$) and increasing for $x \ge 1$ (there $g'(x) \ge 0$). A graph generally consists of several such monotonicity regions.
Strict monotonicity. A differentiable function $f$ with $f'(x) > 0$ for all $x$ is strictly monotonically increasing; if $f'(x) < 0$ for all $x$, it is strictly monotonically decreasing.
The converse of the strict statement fails: the strictly increasing function $f(x) = \tfrac{1}{3}x^3$ has $f'(x) = x^2 \ge 0$, with equality at $x = 0$. So $f' > 0$ everywhere is sufficient but not necessary for strict monotonicity.
The Derivative of an Inverse Function
Strictly monotonic functions are invertible. If $f$ is differentiable with $f'(x) \neq 0$ throughout, then its inverse $f^{-1}$ is differentiable, and differentiating $f^{-1}(f(x)) = x$ with the chain rule gives:
Derivative of the inverse function. $\bigl(f^{-1}\bigr)'(\bar{x}) = \frac{1}{f'(x)}, \qquad \text{where } \bar{x} = f(x).$
For example, from $f(x) = x^2$ (with $x > 0$) and $f'(x) = 2x$ we recover $\bigl(\sqrt{\bar{x}}\bigr)' = \tfrac{1}{2\sqrt{\bar{x}}}$; and from $\exp'(x) = e^x$ we obtain the derivative of the natural logarithm $\ln'(\bar{x}) = \frac{1}{\bar{x}}, \qquad \bar{x} > 0.$
Global Extrema
A location $x_0$ is a global maximum of $f$ if $f(x) \le f(x_0)$ for all $x$, and a global minimum if $f(x) \ge f(x_0)$ for all $x$. Either kind is called a global extremum.
Global extrema of monotone functions. If $f : \lbrack a, b \rbrack \to \mathbb{R}$ is monotone on a closed interval, then the endpoints $a$ and $b$ are its global extrema.
For functions that are not monotone as a whole — the typical case — global extrema may occur in the interior of the interval, and we must also examine local extrema.
Local Extrema
A point $x_0$ is a local maximum of $f$ if there is a surrounding interval $(x_0 - c, x_0 + c)$ with $c > 0$ on which $f(x) \le f(x_0)$, and a local minimum if $f(x) \ge f(x_0)$ there. Either kind is a local extremum, and $(x_0 \mid f(x_0))$ is the corresponding extreme point of the graph.
Necessary Criterion
At a local extremum of a differentiable function, the graph has a horizontal tangent.
Necessary criterion for local extrema. If a differentiable function $f$ has a local extremum at $x_0$, then $f'(x_0) = 0.$
The condition is necessary but not sufficient: for $f(x) = \tfrac{1}{3}x^3$ we have $f'(0) = 0$, yet $x_0 = 0$ is not an extremum, because $f'$ does not change sign there.
Sufficient Criterion via Sign Change
The candidates found from $f'(x_0) = 0$ are genuine extrema precisely when the derivative changes sign there.
- If $f'$ changes sign from $+$ to $-$ at $x_0$, then $x_0$ is a local maximum.
- If $f'$ changes sign from $-$ to $+$ at $x_0$, then $x_0$ is a local minimum.
For the cubic $f(x) = \tfrac{1}{4}x^3 + \tfrac{3}{4}x^2 - \tfrac{3}{2}x - 2$, solving $f'(x) = \tfrac{3}{4}(x^2 + 2x - 2) = 0$ gives the candidates $x_1 = -1 + \sqrt{3}$ and $x_2 = -1 - \sqrt{3}$; checking the sign of the factored derivative shows $x_2$ is a local maximum and $x_1$ a local minimum.
Higher Derivatives
If the derivative $f'$ is itself differentiable, its derivative $(f')'$ is the second derivative $f''$ (read "f two dash"). More generally, the $i$-th derivative $f^{(i)}$ is the derivative of $f^{(i-1)}$, and $f$ is infinitely often differentiable if $f^{(i)}$ exists for all $i \in \mathbb{N}$.
Higher derivatives of polynomials. Every polynomial of degree $n$ can be differentiated infinitely often; its $i$-th derivative is a polynomial of degree $n - i$ for $i \le n$, and vanishes for $i \ge n + 1$.
Not every once-differentiable function can be differentiated repeatedly: $f(x) = x^{5/2}$ ($x \ge 0$) has $f''(x) = \tfrac{15}{4}\sqrt{x}$, but $f''$ is not differentiable at $x = 0$, so $f$ is twice but not three times differentiable.
Sufficient Criterion via the Second Derivative
The second derivative provides a more convenient test than checking sign changes directly.
Sufficient criterion for local extrema. Let $f$ be twice differentiable with $f'(x_0) = 0$. If in addition $f''(x_0) \neq 0$, then $f$ has a local extremum at $x_0$:
- If $f''(x_0) < 0$, then $x_0$ is a local maximum.
- If $f''(x_0) > 0$, then $x_0$ is a local minimum.
For the cubic above, $f''(x) = \tfrac{3}{2}(x + 1)$ gives $f''(x_1) = \tfrac{3\sqrt{3}}{2} > 0$ and $f''(x_2) = -\tfrac{3\sqrt{3}}{2} < 0$, confirming that $x_1$ is a local minimum and $x_2$ a local maximum.
The Higher-Order Test
If $f''(x_0) = 0$ as well, one can sometimes decide using still higher derivatives.
Higher-order criterion. Suppose $f$ is $i$-times differentiable with $f'(x_0) = f''(x_0) = \cdots = f^{(i-1)}(x_0) = 0 \quad\text{and}\quad f^{(i)}(x_0) \neq 0.$ If $i$ is even, then $f$ has a local extremum at $x_0$ — a local maximum if $f^{(i)}(x_0) < 0$, a local minimum if $f^{(i)}(x_0) > 0$. If $i$ is odd, this test does not apply.
For $f(x) = \tfrac{1}{12}x^4$ we have $f'(0) = f''(0) = f'''(0) = 0$ but $f''''(0) = 2 > 0$; since $i = 4$ is even, $x_0 = 0$ is a local minimum. For $f(x) = \tfrac{1}{3}x^3$ the first non-zero derivative at $0$ is $f'''(0) = 2$ with $i = 3$ odd, and indeed $0$ is not an extremum.
Where This Leads
Having located maxima and minima, the next page, Curve Sketching, studies the remaining features of a graph — convexity, inflection points, zeros, symmetry, and asymptotic behaviour — and combines all of these tools to sketch a function completely.