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

Concept: Derivative Sign Analysis – we find critical points where $f'(x)=0$, then check the sign change of $f'(x)$ around each point to classify it as a local maximum, minimum, or neither.

Step 1: Compute the first derivative.

Step 2: Set $f'(x)=0$.

This is the only critical point. …

The key to finding local maxima and minima of a differentiable function is to study where its derivative changes sign. A local maximum occurs where the derivative goes from positive to negative; a local minimum occurs where it goes from negative to positive. If the derivative touches zero but does not cross it — that is, it stays positive on both sides — then the point is not an extremum but a point of inflection with a horizontal tangent.

Let’s apply this logic to the cubic $f(x) = 2x^3 - 6x^2 + 6x + 5$.

1. Compute the first derivative. Differentiate term by term:

Factor out the common factor 6:

2. Find the critical points. Set $f'(x) = 0$:

So $x = 1$ is the only critical point.

3. Analyze the sign of $f'(x)$ around $x = 1$. Since $(x-1)^2 \ge 0$ for all real $x$, and $6 > 0$, we have:

The derivative is zero only at $x = 1$ and positive everywhere else.

4. Interpret the result. …

Method: Confirming a Critical Point Is Genuinely a Local Extremum

A critical point ($f'(x)=0$) is only a candidate for a local maximum or minimum — this method is about the extra check that separates a real extremum from a false alarm.

Steps

Step 1: Find $f'(x)$ and solve $f'(x)=0$

This gives the candidate critical point(s).

Step 2: Examine the sign of $f'$ strictly on both sides of the critical point — not just at it

Do this by testing points immediately to the left and right, or by studying the algebraic form of $f'(x)$ (for example, whether it is a perfect square, which never goes negative).

Step 3: Interpret a "no sign change" result correctly …

Common Mistakes

Mistake 1: Concluding $x=1$ is an extremum just because $f'(1)=0$

Why it's wrong: a zero derivative only makes a point a candidate; it becomes an actual local max or min only if the sign of $f'$ changes across it. Here $f'(x)=6(x-1)^2$ never goes negative, so despite $f'(1)=0$, the function keeps increasing straight through $x=1$. Correct approach: always test the sign of $f'$ on both sides of a critical point before declaring it a maximum or minimum — never stop at "the derivative is zero here."

Mistake 2: Missing that a repeated (squared) factor signals "no sign change" …