Q.The function $f(x) = 2x^3 - 3x^2 - 12x + 4$ has:
(A) two points of local maximum
(B) two points of local minimum
(C) one maxima and one minima
(D) no maxima or minima

Concept: Derivative Sign Analysis — local extrema occur where $f'(x)=0$ and the sign of $f'(x)$ changes.

1. Compute the derivative:

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

2. Test sign of $f'(x)$ on intervals:
   • For $x < -1$, say $x=-2$: $f'(-2)=6(-4)(-1)=24 > 0$ → increasing.
   • For $-1 < x < 2$, say $x=0$: $f'(0)=6(-2)(1)=-12 < 0$ → decreasing. …

To decide whether a function has local maxima or minima, we look at where its derivative changes sign. A local maximum occurs when $f'$ goes from positive to negative; a local minimum when $f'$ goes from negative to positive. For a polynomial like this cubic, the derivative is a quadratic — so it can have at most two real roots, and the sign pattern around those roots tells us everything.

Let’s work through it.

1. Find the derivative. $f(x) = 2x^3 - 3x^2 - 12x + 4$ Differentiating term by term:

2. Factor the derivative to find critical points. Factor out the common 6:

The quadratic factorises:

So

Setting $f'(x) = 0$ gives the critical points:

3. Analyse the sign of $f'$ on the number line.

   The factors $(x+1)$ and $(x-2)$ are linear, so the sign of $f'$ changes at each root. Since the leading coefficient of $f'$ is positive ($6 > 0$), the quadratic opens upward — meaning $f'$ is positive outside the interval between the roots and negative inside it.

   Let’s check explicitly:

   • For $x < -1$: both $(x+1)$ and $(x-2)$ are negative, product positive → $f'(x) > 0$
   • For $-1 < x < 2$: $(x+1)$ positive, $(x-2)$ negative → product negative → $f'(x) < 0$
   • For $x > 2$: both factors positive → $f'(x) > 0$

   So the sign pattern is:

4. Interpret the sign changes.
   • At $x = -1$: $f'$ goes from positive to negative → local maximum. …

Method: Locating and Counting Local Maxima/Minima via the First Derivative Test

Use this whenever a question asks how many local maxima and/or minima a polynomial (typically cubic or higher) has, without specifying an interval.

Steps

Step 1: Differentiate the function to get $f'(x)$.

For a cubic, this gives a quadratic derivative, which can have $0$, $1$, or $2$ real roots.

Step 2: Factor $f'(x)$ (or use the quadratic formula) to find the critical points.

Solve $f'(x)=0$; a positive discriminant means two distinct real critical points, which is exactly when both a local maximum and a local minimum can occur.

Step 3: Determine the sign of $f'(x)$ on each of the three resulting intervals. …

Common Mistakes

Mistake 1: Assuming a cubic can have only one turning point.

Why it's wrong: a cubic's derivative is a quadratic, which can have two distinct real roots — so the cubic itself can have both a local maximum and a local minimum, not just one turning point. Correct approach: always check the discriminant of the derivative; a positive discriminant means two genuine critical points to classify.

Mistake 2: Concluding "maximum" or "minimum" just from the value $f'(x)=0$ at the critical point itself.

Why it's wrong: $f'(x)=0$ exactly at a critical point tells you nothing about whether it's a peak or valley — the classification depends on how the sign changes on either side, not the value at the point. Correct approach: always test the sign of $f'$ strictly on both sides of each critical point (or use the second derivative test) before labelling it max or min. …