Maxima and minima — first vs second derivative test selection

Question

Find the local maxima and minima of $f(x) = 2x^3 - 9x^2 + 12x - 3$ using both the first and second derivative tests. When should we prefer one test over the other?

(CBSE 12 + JEE Main pattern)

Solution — Step by Step

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)

Critical points: $x = 1$ and $x = 2$ .

f''(x) = 12x - 18

At $x = 1$ : $f''(1) = 12 - 18 = -6 < 0$ → Local maximum

At $x = 2$ : $f''(2) = 24 - 18 = 6 > 0$ → Local minimum

$f(1) = 2 - 9 + 12 - 3 = 2$ (local max value)

$f(2) = 16 - 36 + 24 - 3 = 1$ (local min value)

Check sign changes of $f'(x)$ :

Interval	$f'(x)$	Behaviour
$x < 1$	$+$	Increasing
$1 < x < 2$	$-$	Decreasing
$x > 2$	$+$	Increasing

At $x = 1$ : $f'$ changes from $+$ to $-$ → Local maximum ✓

At $x = 2$ : $f'$ changes from $-$ to $+$ → Local minimum ✓

flowchart TD
    A["Find f'(x) = 0 → critical points"] --> B{"Which test to use?"}
    B -->|"Second derivative test"| C["Compute f''(x) at critical point"]
    C --> D{"f''(c) < 0?"}
    D -->|"Yes"| E["Local Maximum"]
    D -->|"No, f''(c) > 0"| F["Local Minimum"]
    D -->|"f''(c) = 0"| G["Test FAILS — use first derivative test"]
    B -->|"First derivative test"| H["Check sign change of f'(x)"]
    H --> I{"+ to - change?"}
    I -->|"Yes"| E
    I -->|"No, - to +"| F
    I -->|"No sign change"| J["Neither max nor min (inflection point)"]

Why This Works

The second derivative test uses concavity. If $f''(c) < 0$ , the curve is concave downward at $c$ — like the top of a hill — so it is a local maximum. If $f''(c) > 0$ , the curve is concave upward — like the bottom of a valley — so it is a local minimum.

The first derivative test uses the sign change of $f'$ . If $f'$ goes from positive to negative at $c$ , the function was rising and then started falling — $c$ is a peak. If $f'$ goes from negative to positive, it was falling and then started rising — $c$ is a valley.

When to prefer which: Use the second derivative test when $f''$ is easy to compute. Use the first derivative test when $f''(c) = 0$ (second derivative test fails) or when $f''$ is messy to calculate.

Alternative Method — Higher Derivative Test

When $f''(c) = 0$ , we go to higher derivatives. Find the first non-zero derivative at $c$ . If the $n$ -th derivative is the first non-zero one:

If $n$ is even and $f^{(n)}(c) > 0$ → local minimum
If $n$ is even and $f^{(n)}(c) < 0$ → local maximum
If $n$ is odd → inflection point (neither max nor min)

For JEE optimization word problems (“find the maximum area,” “minimize the cost”), always check endpoints too, not just critical points. A function on a closed interval $[a, b]$ attains its absolute maximum and minimum at either a critical point or an endpoint. Students who only check critical points miss the actual answer when it occurs at a boundary.

Common Mistake

When $f''(c) = 0$ , students conclude “neither max nor min.” That is not always true — $f''(c) = 0$ simply means the second derivative test is inconclusive. The point COULD still be a max or min; you need to check using the first derivative test or higher-order derivatives. For example, $f(x) = x^4$ has $f''(0) = 0$ , but $x = 0$ IS a local minimum (confirmed by $f^{(4)}(0) = 24 > 0$ ).

Question

Solution — Step by Step

Why This Works

Alternative Method — Higher Derivative Test

Common Mistake

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