Find maxima and minima of f(x) = 2x³ - 3x² - 12x + 5 using second derivative test

Question

Find the local maxima and minima of $f(x) = 2x^3 - 3x^2 - 12x + 5$ using the second derivative test.

(CBSE 2022, 4 marks)

Solution — Step by Step

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

Setting $f'(x) = 0$ : $x = 2$ or $x = -1$ .

These are the critical points.

f''(x) = 12x - 6

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

Since $f''(-1) < 0$ , $x = -1$ is a local maximum.

At $x = 2$ : $f''(2) = 12(2) - 6 = 18 > 0$

Since $f''(2) > 0$ , $x = 2$ is a local minimum.

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$

$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$

\boxed{\text{Local maximum} = 12 \text{ at } x = -1}

\boxed{\text{Local minimum} = -15 \text{ at } x = 2}

Why This Works

The first derivative tells us where the slope is zero — potential turning points. The second derivative tells us the concavity at those points. If $f'' > 0$ , the curve is concave up (cup-shaped), so the critical point is a minimum. If $f'' < 0$ , the curve is concave down (cap-shaped), so it’s a maximum.

For this cubic, the graph rises to a peak at $x = -1$ (value 12), then drops to a trough at $x = 2$ (value $-15$ ), then rises again. The difference $12 - (-15) = 27$ gives the “swing” of the function between its extrema.

Alternative Method — First derivative test

Check the sign of $f'(x)$ around each critical point:

For $x = -1$ : $f'(-2) = 6(4+2-2) = 24 > 0$ and $f'(0) = -12 < 0$ . Sign changes from $+$ to $-$ → local maximum.

For $x = 2$ : $f'(1) = 6(1-1-2) = -12 < 0$ and $f'(3) = 6(9-3-2) = 24 > 0$ . Sign changes from $-$ to $+$ → local minimum.

The first derivative test works when $f'' = 0$ (second derivative test is inconclusive). But for CBSE boards, use whichever test the question specifies. If it says “using second derivative test,” you must use it — using the first derivative test instead will lose marks, even if the answer is correct.

Common Mistake

Students sometimes confuse the sign convention: $f'' > 0$ means minimum (not maximum). A positive second derivative means the curve bends upward — like the bottom of a bowl. A negative second derivative means it bends downward — like the top of a hill. If you mix this up, you’ll swap the maxima and minima labels.

Question

Solution — Step by Step

Why This Works

Alternative Method — First derivative test

Common Mistake

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