Maxima and Minima — First vs Second Derivative Test Selection

Question

When finding maxima and minima, how do we choose between the first derivative test and the second derivative test?

Solution — Step by Step

For both tests, start by finding where $f'(x) = 0$ or $f'(x)$ is undefined. These are the critical points.

Example: $f(x) = x^3 - 3x + 2$

f'(x) = 3x^2 - 3 = 0 \implies x = \pm 1

Look at the sign of $f'(x)$ on either side of the critical point:

$f'$ changes from $+$ to $-$ at $x = c$ $\implies$ local maximum at $c$
$f'$ changes from $-$ to $+$ at $x = c$ $\implies$ local minimum at $c$
$f'$ does NOT change sign $\implies$ neither (inflection point)

For $x = -1$ : $f'(-2) = 9 > 0$ , $f'(0) = -3 < 0$ . Sign changes $+$ to $-$ : local maximum. For $x = 1$ : $f'(0) = -3 < 0$ , $f'(2) = 9 > 0$ . Sign changes $-$ to $+$ : local minimum.

Compute $f''(x)$ and evaluate at the critical point:

$f''(c) < 0 \implies$ local maximum (concave down)
$f''(c) > 0 \implies$ local minimum (concave up)
$f''(c) = 0 \implies$ test fails (use first derivative test instead)

For our example: $f''(x) = 6x$ .

$f''(-1) = -6 < 0$ : local maximum at $x = -1$
$f''(1) = 6 > 0$ : local minimum at $x = 1$

graph TD
    A[Found critical point c where f prime c = 0] --> B{Is f double prime easy to compute?}
    B -->|Yes| C[Compute f double prime at c]
    C --> D{f double prime at c = 0?}
    D -->|No, negative| E[Local maximum]
    D -->|No, positive| F[Local minimum]
    D -->|Yes, equals 0| G[Second derivative test FAILS]
    G --> H[Must use first derivative test]
    B -->|No, or f prime has factors| I[Use first derivative test directly]
    I --> J[Check sign change of f prime around c]

Rule of thumb: Use the second derivative test when $f''(x)$ is simple to compute. Use the first derivative test when $f''(c) = 0$ or when $f'(x)$ is already factored.

Why This Works

The first derivative test looks at the “direction change” — if the function was going up and starts going down, we have a peak. The second derivative test looks at “curvature” — if the curve is concave down at a flat point, it must be a peak.

Both tests give the same answer, but the second derivative test is faster (one computation vs. checking two intervals). Its weakness is that it fails when $f''(c) = 0$ , which happens at inflection points.

For JEE Main, the second derivative test is usually faster and sufficient. But for functions like $f(x) = x^4$ at $x = 0$ (where $f''(0) = 0$ ), you must fall back to the first derivative test. Knowing when to switch is the real skill.

Alternative Method

For optimization word problems (maximise area, minimise cost), after finding the critical point, you often do not need to test at all — if there is only one critical point in the valid domain and the function goes to infinity (or zero) at the boundaries, the critical point must be the desired extremum.

Common Mistake

When $f''(c) = 0$ , students often conclude “no maximum or minimum.” This is wrong. $f''(c) = 0$ means the test is inconclusive — the point could still be a max, min, or neither. For example, $f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$ , but $x = 0$ IS a local minimum (check using first derivative test: $f'$ changes from $-$ to $+$ ). Always switch to the first derivative test when the second fails.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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