Maxima and Minima — First and Second Derivative Tests

The Idea of Maximum and Minimum

This is the heart of the chapter: finding the turning points of a graph — the peaks and the valleys. Real problems lead here naturally: how many trees per acre maximise an orchard's profit? What is the maximum height a thrown ball reaches? What is the nearest a helicopter on the curve $y = x^2 + 7$ comes to a soldier on the ground?

Let $f$ be a function defined on an interval $I$. Then:

• $f$ has a maximum value in $I$ if there is a point $c \in I$ with $f(c) > f(x)$ for all $x \in I$. The number $f(c)$ is the maximum value, and $c$ is a point of maximum value.
• $f$ has a minimum value in $I$ if there is a $c \in I$ with $f(c) < f(x)$ for all $x \in I$.
• An extreme value is either of these; $c$ is then an extreme point.

Three instructive quick cases:

1. $f(x) = x^2$ on $\mathbb{R}$: minimum value 0 at $x = 0$, but no maximum — the parabola climbs forever.
2. $f(x) = |x|$ on $\mathbb{R}$: minimum value 0 at $x = 0$ — even though $f$ is not differentiable there. Extrema can live at corners!
3. $f(x) = x$ on the open interval $(0, 1)$: neither a maximum nor a minimum — any candidate point has another point beyond it, and the endpoints are not included.

Key Point: Two facts worth memorising: every monotonic function attains its max/min at the endpoints of its domain, and every continuous function on a closed interval $[a, b]$ has both an absolute maximum and an absolute minimum (Theorem 5).

[JEE Tip] Open vs closed intervals is exactly where MCQ setters lay traps. On $(0, 1)$ the function $x$ has no extrema; on $[0, 1]$ it has both. Check the brackets before answering.

Local Maxima, Local Minima and Critical Points

Most graphs are not monotonic — they rise, fall, rise again. At each turning point the graph has a little hill (local maximum) or a little valley (local minimum).

Definition. Let $c$ be an interior point of the domain of $f$. Then:

• $c$ is a point of local maxima if there is an $h > 0$ such that $f(c) \geq f(x)$ for all $x \in (c-h, c+h)$. The value $f(c)$ is the local maximum value.
• $c$ is a point of local minima if there is an $h > 0$ such that $f(c) \leq f(x)$ for all $x \in (c-h, c+h)$.

"Local" means: compared only with nearby points, not the whole domain.

Theorem 2. If $f$ has a local maximum or local minimum at $x = c$, then either $f'(c) = 0$ or $f$ is not differentiable at $c$.

This motivates the key definition: a point $c$ where either $f'(c) = 0$ or $f$ is not differentiable is called a critical point of $f$. Critical points are the only candidates for local extrema.

Key Point: The converse of Theorem 2 is FALSE. For $f(x) = x^3$, $f'(0) = 0$, yet $x = 0$ is neither a local max nor a local min — the curve just flattens momentarily and keeps climbing. $f'(c) = 0$ makes $c$ a candidate, not a winner.

[JEE Important] "Number of critical points" questions must count BOTH kinds: zeros of $f'$ AND points of non-differentiability (corners, cusps). Forgetting the second kind is a classic JEE error.

The First Derivative Test

How do we decide whether a critical point is a hill, a valley, or neither? Watch the sign of $f'$ as $x$ passes through $c$.

Theorem 3 (First Derivative Test). Let $f$ be continuous at a critical point $c$ in an open interval $I$. Then:

1. If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$, then $c$ is a point of local maxima. (Rising, then falling — a hilltop.)
2. If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$, then $c$ is a point of local minima. (Falling, then rising — a valley floor.)
3. If $f'(x)$ does not change sign through $c$, then $c$ is neither — it is a point of inflection.

Worked pattern: $f(x) = x^3 - 3x + 3$ gives $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$, so the critical points are $x = \pm 1$.

• Through $x = -1$: $f'$ goes $(+) \to (-)$ — local maximum, value $f(-1) = 5$.
• Through $x = 1$: $f'$ goes $(-) \to (+)$ — local minimum, value $f(1) = 1$.

Key Point: Test the sign of $f'$ at convenient points just left and just right of each critical point (e.g. at $0.9$ and $1.1$ for $c = 1$). A tiny sign table earns full method marks.

[Board Important] Always report both the point ($x = c$) and the value ($f(c)$) — questions ask for "local maximum value", and stopping at $x = c$ loses a mark.

The Second Derivative Test

Often quicker: differentiate twice and check concavity at the critical point.

Theorem 4 (Second Derivative Test). Let $f$ be twice differentiable at $c$ with $f'(c) = 0$. Then:

1. If $f''(c) < 0$: $x = c$ is a point of local maxima (curve bends downwards — a dome).
2. If $f''(c) > 0$: $x = c$ is a point of local minima (curve bends upwards — a cup).
3. If $f''(c) = 0$: the test fails. Go back to the first derivative test and check the sign change.

Worked pattern: $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

$$f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2), \qquad f''(x) = 36x^2 + 24x - 24$$

Critical points: $x = 0, 1, -2$. Now $f''(0) = -24 < 0$, $f''(1) = 36 > 0$, $f''(-2) = 72 > 0$. So $x = 0$ is a local maximum with $f(0) = 12$, while $x = 1$ and $x = -2$ are local minima with $f(1) = 7$ and $f(-2) = -20$.

When the test fails: $f(x) = 2x^3 - 6x^2 + 6x + 5$ has $f'(x) = 6(x-1)^2$ and $f''(x) = 12(x-1)$, so $f'(1) = 0$ AND $f''(1) = 0$ — no verdict. First derivative test: $f'(x) = 6(x-1)^2 \geq 0$ on both sides (no sign change), so $x = 1$ is a point of inflexion.

Key Point: Second derivative test = fast but can fail. First derivative test = always works (for continuous $f$). When $f''(c) = 0$, fall back — never conclude "no extremum" just because the second test is silent.

[JEE Tip] For functions with corners (like $3 + |x|$), the second derivative test is unusable at the corner — only the first derivative test applies there. Sign of $f'$: negative left of 0, positive right of 0 $\Rightarrow$ local minimum value $3$ at $x = 0$.

Solved Examples

Example 1: A parabola's extremes

Find the maximum and the minimum values, if any, of $f(x) = x^2$, $x \in \mathbb{R}$.

Solution:

1. $f(x) = x^2 \geq 0$ for all $x$, and $f(0) = 0$.
2. So the minimum value is 0, attained at the point of minimum value $x = 0$.
3. As $x \to \pm\infty$, $f(x) \to \infty$: no maximum value exists on $\mathbb{R}$.

Final Answer: Minimum value 0 at $x = 0$; no maximum value.

Takeaway: If the domain were restricted to $[-2, 1]$, a maximum WOULD appear: $f(-2) = 4$. Domains decide extrema.

Example 2: A corner can hold a minimum

Find the maximum and minimum values, if any, of $f(x) = |x|$, $x \in \mathbb{R}$.

Solution:

1. $|x| \geq 0$ for all $x$ and $|0| = 0$: minimum value 0 at $x = 0$.
2. The graph rises without bound on both sides: no maximum value.
3. Note $f$ is NOT differentiable at $x = 0$ — yet the minimum lives exactly there.

Final Answer: Minimum value 0 at $x = 0$; no maximum.

Takeaway: Extreme values can occur at points of non-differentiability. Critical points include corners, not just flat tangents.

Example 3: An open interval with no extremes

Find the maximum and minimum values, if any, of $f(x) = x$, $x \in (0, 1)$.

Solution:

1. $f$ is strictly increasing on $(0, 1)$, so extremes could only be near the ends.
2. But for any $x_0 \in (0, 1)$, the point $\frac{x_0}{2}$ is smaller still, and $\frac{x_1 + 1}{2}$ exceeds any candidate $x_1$: no smallest or largest value is ever attained.

Final Answer: Neither a maximum nor a minimum value exists on $(0,1)$.

Takeaway: On the closed interval $[0, 1]$ the same function has minimum 0 and maximum 1 — endpoints matter. A continuous function on a closed interval always attains both extremes.

Example 4: First derivative test in action

Find all points of local maxima and local minima of $f(x) = x^3 - 3x + 3$.

Solution:

1. Differentiate and factorise: $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$; critical points $x = \pm 1$.
2. Through $x = 1$: left of 1 (say 0.9): $f' < 0$; right of 1 (say 1.1): $f' > 0$. Sign change $(-) \to (+)$ $\Rightarrow$ local minimum, value $f(1) = 1 - 3 + 3 = 1$.
3. Through $x = -1$: left (say $-1.1$): $f' > 0$; right (say $-0.9$): $f' < 0$. Sign change $(+) \to (-)$ $\Rightarrow$ local maximum, value $f(-1) = -1 + 3 + 3 = 5$.

Final Answer: Local maximum value 5 at $x = -1$; local minimum value 1 at $x = 1$.

Takeaway: For a cubic with positive leading coefficient, the left critical point is always the local max and the right one the local min.

Example 5: A point of inflexion

Find all the points of local maxima and local minima of $f(x) = 2x^3 - 6x^2 + 6x + 5$.

Solution:

1. Differentiate: $f'(x) = 6x^2 - 12x + 6 = 6(x - 1)^2$; the only critical point is $x = 1$.
2. Check the sign change: $f'(x) = 6(x-1)^2 \geq 0$ on BOTH sides of 1 — no sign change.
3. Conclude: by the first derivative test, $x = 1$ is neither a local max nor a local min: it is a point of inflexion.
4. Cross-check with the second derivative: $f''(x) = 12(x - 1)$ gives $f''(1) = 0$ — the second derivative test fails here, as expected.

Final Answer: No local maxima or minima; $x = 1$ is a point of inflexion.

Takeaway: A perfect-square derivative never changes sign — the graph pauses (horizontal tangent) but keeps climbing.

Example 6: A corner minimum

Find the local minimum value of $f(x) = 3 + |x|$, $x \in \mathbb{R}$.

Solution:

1. $f$ is not differentiable at $x = 0$, so the second derivative test is unusable; $x = 0$ is still a critical point.
2. Left of 0: $f(x) = 3 - x$, $f'(x) = -1 < 0$. Right of 0: $f(x) = 3 + x$, $f'(x) = 1 > 0$.
3. Sign change $(-) \to (+)$ $\Rightarrow$ local minimum at $x = 0$.

Final Answer: Local minimum value $f(0) = 3$.

Takeaway: The first derivative test handles corners gracefully — it only needs signs on either side, not a derivative at the point itself.

Example 7: Second derivative test on a quartic

Find the local maximum and local minimum values of $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

Solution:

1. Differentiate and factorise: $f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2)$; critical points $x = 0, 1, -2$.
2. Second derivative: $f''(x) = 36x^2 + 24x - 24$.
3. Evaluate: $f''(0) = -24 < 0 \Rightarrow$ local max at 0; $f''(1) = 36 > 0 \Rightarrow$ local min at 1; $f''(-2) = 144 - 48 - 24 = 72 > 0 \Rightarrow$ local min at $-2$.
4. Values: $f(0) = 12$; $f(1) = 3 + 4 - 12 + 12 = 7$; $f(-2) = 48 - 32 - 48 + 12 = -20$.

Final Answer: Local maximum value 12 at $x = 0$; local minimum values 7 at $x = 1$ and $-20$ at $x = -2$.

Takeaway: With three critical points, the second derivative test is much faster than three sign tables — that's exactly when to prefer it.

Example 8: The standard cubic $x^3 - 3x$

Find the local maxima and minima of $g(x) = x^3 - 3x$, and the corresponding values.

Solution:

1. $g'(x) = 3x^2 - 3 = 3(x-1)(x+1)$: critical points $x = \pm 1$.
2. $g''(x) = 6x$: $g''(-1) = -6 < 0 \Rightarrow$ local max at $x = -1$; $g''(1) = 6 > 0 \Rightarrow$ local min at $x = 1$.
3. Values: $g(-1) = -1 + 3 = 2$; $g(1) = 1 - 3 = -2$.

Final Answer: Local maximum value 2 at $x = -1$; local minimum value $-2$ at $x = 1$.

Takeaway: Notice the local max value (2) is smaller than values of $g$ far away (e.g. $g(10) = 970$) — "local" really does mean local.

Example 9: A trigonometric local maximum

Find the local maximum of $h(x) = \sin x + \cos x$, $0 < x < \frac{\pi}{2}$.

Solution:

1. $h'(x) = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}$ (the only solution in the domain).
2. $h''(x) = -\sin x - \cos x$, so $h''\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0$: local maximum.
3. Value: $h\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.

Final Answer: Local maximum value $\sqrt{2}$ at $x = \frac{\pi}{4}$.

Takeaway: $\sqrt{2}$ is also the global maximum of $\sin x + \cos x$ over all reals — remember $-\sqrt{a^2+b^2} \leq a\sin x + b\cos x \leq \sqrt{a^2+b^2}$.

Example 10: A cubic with two turning points

Find the local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$.

Solution:

1. $f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$: critical points $x = 1, 3$.
2. $f''(x) = 6x - 12$: $f''(1) = -6 < 0$ (local max); $f''(3) = 6 > 0$ (local min).
3. Values: $f(1) = 1 - 6 + 9 + 15 = 19$; $f(3) = 27 - 54 + 27 + 15 = 15$.

Final Answer: Local maximum value 19 at $x = 1$; local minimum value 15 at $x = 3$.

Takeaway: This is the single most common Board format for a 3-mark maxima-minima question. Practise it until it takes under three minutes.

Example 11: A rational function minimum

Find the local minimum of $g(x) = \frac{x}{2} + \frac{2}{x}$, $x > 0$.

Solution:

1. $g'(x) = \frac{1}{2} - \frac{2}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$ (taking $x > 0$).
2. $g''(x) = \frac{4}{x^3}$, so $g''(2) = \frac{1}{2} > 0$: local minimum.
3. Value: $g(2) = 1 + 1 = 2$.

Final Answer: Local minimum value 2 at $x = 2$.

Takeaway: This matches the AM-GM inequality: $\frac{x}{2} + \frac{2}{x} \geq 2\sqrt{\frac{x}{2}\cdot\frac{2}{x}} = 2$ — a great way to verify derivative answers instantly.

Example 12: A square-root product

Find the local maximum of $f(x) = x\sqrt{1 - x}$, $0 < x < 1$.

Solution:

1. Differentiate (product + chain rule): $f'(x) = \sqrt{1-x} + x \cdot \frac{-1}{2\sqrt{1-x}} = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$.
2. Critical point: $f'(x) = 0 \Rightarrow x = \frac{2}{3}$ (inside the domain).
3. Sign change: $f' > 0$ for $x < \frac{2}{3}$ and $f' < 0$ for $x > \frac{2}{3}$: local maximum.
4. Value: $f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.

Final Answer: Local maximum value $\frac{2\sqrt{3}}{9}$ at $x = \frac{2}{3}$.

Takeaway: Combine the fraction before hunting for zeros — a single tidy numerator makes both the roots and the sign analysis obvious.

Example 13: Functions with no extrema at all

Prove that $f(x) = e^x$, $g(x) = \log x$ and $h(x) = x^3 + x^2 + x + 1$ have no maxima or minima.

Solution:

1. $e^x$: $f'(x) = e^x > 0$ for every $x$ — never zero, always differentiable, so no critical points and hence no extrema.
2. $\log x$: $g'(x) = \frac{1}{x} > 0$ on the domain $(0, \infty)$ — same conclusion.
3. $h(x)$: $h'(x) = 3x^2 + 2x + 1$ has discriminant $4 - 12 = -8 < 0$, so $h'(x) > 0$ for all $x$ — no real critical points, no extrema.

Final Answer: All three functions are strictly increasing with no critical points, hence no maxima or minima.

Takeaway: No critical points $\Rightarrow$ no local extrema. A negative discriminant of a quadratic derivative settles it in one line.