Maxima and minima

A local maximum of $f$ at $c$ means $f(c) \ge f(x)$ for all $x$ in a small neighbourhood of $c$. A local minimum is the same with the inequality reversed. The corresponding absolute (or global) extrema are the largest/smallest values of $f$ on its entire domain. Derivatives find both.

Definitions

• $f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in some open interval containing $c$.
• $f$ has a local minimum at $c$ if $f(c) \le f(x)$ for all $x$ in some open interval containing $c$.
• The point $c$ is a critical point if $f'(c) = 0$ or $f'(c)$ is undefined.

Every interior local extremum of a differentiable $f$ is a critical point (Fermat's theorem). The converse is false: critical points may be neither.

First derivative test

At a critical point $c$:

• If $f'$ changes from $+$ to $-$ as $x$ crosses $c$ left to right: local maximum.
• If $f'$ changes from $-$ to $+$: local minimum.
• If $f'$ keeps the same sign: neither (an inflection-like point).

Second derivative test

At a critical point $c$ where $f''(c)$ exists:

• $f''(c) > 0 \Rightarrow$ local minimum.
• $f''(c) < 0 \Rightarrow$ local maximum.
• $f''(c) = 0 \Rightarrow$ test fails; use first derivative test or higher-order analysis.

Absolute extrema on a closed interval

To find the absolute max/min of a continuous $f$ on $[a, b]$:

1. Find critical points in $(a, b)$.
2. Evaluate $f$ at all critical points and at the endpoints $a, b$.
3. The largest value is the absolute maximum; smallest is the absolute minimum.

This is the closed-interval method and is the most common board examination technique.

Worked example

$f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$.

$f'(x) = 3x^2 - 6x = 3x(x - 2)$. Critical points: $x = 0, 2$.

Values: $f(-1) = -1 - 3 + 1 = -3$; $f(0) = 1$; $f(2) = 8 - 12 + 1 = -3$; $f(4) = 64 - 48 + 1 = 17$.

Absolute max: $17$ at $x = 4$. Absolute min: $-3$ at $x = -1$ and $x = 2$.

Worked examples

Example 1. Find local extrema of $f(x) = x^3 - 6x^2 + 9x + 7$.

$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$.

Critical points: $x = 1, 3$. $f''(x) = 6x - 12$. $f''(1) = -6 < 0$, so $x = 1$ is local max with $f(1) = 11$. $f''(3) = 6 > 0$, so $x = 3$ is local min with $f(3) = 7$.

Example 2. Find local extrema of $f(x) = x^4 - 4x^3 + 4x^2 + 1$.

$f'(x) = 4x^3 - 12x^2 + 8x = 4x(x - 1)(x - 2)$. Critical points $0, 1, 2$.

$f''(x) = 12 x^2 - 24 x + 8$. $f''(0) = 8 > 0$, local min, $f(0) = 1$. $f''(1) = -4 < 0$, local max, $f(1) = 2$. $f''(2) = 8 > 0$, local min, $f(2) = 1$.

Example 3. Find absolute max and min of $f(x) = \sin x + \cos x$ on $[0, \pi]$.

$f'(x) = \cos x - \sin x$. Critical: $\cos x = \sin x$, i.e., $x = \pi/4$.

$f(0) = 1, f(\pi/4) = \sqrt 2 \approx 1.414, f(\pi) = -1$.

Max: $\sqrt 2$ at $\pi/4$. Min: $-1$ at $\pi$.

Example 4. Show that $f(x) = x^3$ has no local extremum at $x = 0$.

$f'(0) = 0$ (critical point), but $f'(x) = 3x^2 \ge 0$ with equality only at $0$. So $f$ is increasing on both sides , no extremum (an inflection point).

Example 5. $f(x) = x e^x$. Find local extrema.

$f'(x) = e^x + x e^x = e^x(1 + x)$. Critical at $x = -1$. $f''(x) = e^x + (1 + x) e^x = e^x(2 + x)$. $f''(-1) = e^{-1}(1) > 0$. Local min, value $-e^{-1}$.

Example 6. Find the absolute extrema of $f(x) = (x - 1)^{2/3}$ on $[0, 2]$.

$f'(x) = (2/3)(x - 1)^{-1/3}$. Undefined at $x = 1$ (cusp), zero nowhere.

Values: $f(0) = 1, f(1) = 0, f(2) = 1$. Absolute max $1$ at $x = 0$ and $x = 2$; absolute min $0$ at $x = 1$.

Try it yourself

1. Find local extrema of $f(x) = x^3 - 3x + 2$.
2. Find local extrema of $f(x) = x^4 - 8 x^2 + 16$.
3. Find absolute max of $f(x) = x^3 - 12 x + 5$ on $[-3, 3]$.
4. Find $a$ such that $f(x) = x^2 + a/x$ has a local minimum at $x = 2$.
5. Find local extrema of $f(x) = \sin x + (1/2)\cos 2x$ on $[0, \pi/2]$.
6. Show that the function $f(x) = x \log x$ has a minimum at $x = 1/e$.
7. Find absolute max/min of $f(x) = x + \sin 2x$ on $[0, 2\pi]$.
8. Find the local extrema of $f(x) = x^{5/3} - 5 x^{2/3}$.
9. Show $f(x) = \tan x - 4x$ is strictly decreasing on $(-\pi/3, \pi/3)$.
10. Find local extrema of $f(x) = x \sqrt{1 - x}$ on $[0, 1]$.
11. Find absolute max of $f(x) = x^2 e^{-x}$ on $[0, 5]$.
12. Find local extrema of $f(x) = \sin x - \sin 2x$ on $[0, 2\pi]$.
13. Find the points where $f(x) = (x - 2)^4 (x + 1)^3$ has local extrema.
14. Use the second derivative test on $f(x) = x^3 - 3 x^2 - 9 x + 5$.

Pitfalls / Tricks

• Critical points include where $f'$ is undefined, not just where $f' = 0$.
• Always check endpoint values when finding absolute extrema on a closed interval.
• The second derivative test is inconclusive at $f''(c) = 0$. Fall back to the first derivative test.
• A "saddle" (inflection point) has $f'(c) = 0$ but no sign change.
• Some problems require the global maximum or minimum , be explicit about whether your answer is local or absolute.

Next, the most JEE-heavy section: applied optimisation.