Q.Find the intervals in which the function $f$ given by $f(x) = 4x^3 - 6x^2 - 72x + 30$ is

Concept: Monotonic Function Analysis — A function increases where $f'(x) > 0$ and decreases where $f'(x) < 0$.

Step 1: Differentiate

$f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6)$.

Step 2: Factor the quadratic

$x^2 - x - 6 = (x - 3)(x + 2)$.

So $f'(x) = 12(x - 3)(x + 2)$.

Step 3: Find critical points

Set $f'(x) = 0$ → $x = -2$ and $x = 3$. These split the real line into three intervals: $(-\infty, -2)$, $(-2, 3)$, $(3, \infty)$.

Step 4: Test sign of $f'(x)$ in each interval

• For $x < -2$: $(x-3)$ negative, $(x+2)$ negative → product positive → $f'(x) > 0$ → increasing.
• For $-2 < x < 3$: $(x-3)$ negative, $(x+2)$ positive → product negative → $f'(x) < 0$ → decreasing. …

The key idea is simple: a function increases where its derivative is positive and decreases where its derivative is negative. For a polynomial like this, the derivative is a quadratic, so its sign changes only at its roots. Once we find those roots, we just check one point in each interval.

Let’s go step by step.

1. Find the derivative.

   $f(x) = 4x^3 - 6x^2 - 72x + 30$

   Differentiate term by term:

   $f'(x) = 12x^2 - 12x - 72$

2. Factor the derivative to find critical points.

   First, take out the common factor 12:

   $f'(x) = 12(x^2 - x - 6)$

   Factor the quadratic:

   $x^2 - x - 6 = (x - 3)(x + 2)$

   So $f'(x) = 12(x - 3)(x + 2)$

3. Set $f'(x) = 0$ to find where the derivative changes sign.

   $12(x - 3)(x + 2) = 0$ gives $x = 3$ and $x = -2$.

   These are the only points where $f'(x)$ could change sign.

4. Divide the real line into intervals using these points.

   The three intervals are:

   $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

5. Test the sign of $f'(x)$ in each interval.

   Pick a convenient test point in each interval and plug into $f'(x) = 12(x - 3)(x + 2)$.

   • For $(-\infty, -2)$, choose $x = -3$:

     $f'(-3) = 12(-3 - 3)(-3 + 2) = 12(-6)(-1) = 72 > 0$

     So $f$ is increasing on $(-\infty, -2)$.

   • For $(-2, 3)$, choose $x = 0$:

     $f'(0) = 12(0 - 3)(0 + 2) = 12(-3)(2) = -72 < 0$

     So $f$ is decreasing on $(-2, 3)$.

   • For $(3, \infty)$, choose $x = 4$: …

Method: Finding Intervals of Increase/Decrease by Factoring the Derivative

When $f'(x)$ is a quadratic (from differentiating a cubic), factoring it is usually faster and less error-prone than the quadratic formula, and it lets you read the sign of each interval directly from the sign of the factors.

Steps

Step 1: Differentiate and pull out any common numerical factor.

Factoring out a positive common constant first doesn't affect the sign, but keeps the numbers small for factoring.

Step 2: Factor the remaining quadratic into two linear factors.

Look for two numbers that multiply to the constant term and add to the coefficient of $x$.

Step 3: Locate the critical points from each factor being zero, and split the real line there. …

Common Mistakes

Mistake 1: Forgetting to factor out the common numerical factor before finding the roots of the quadratic.

Why it's wrong: trying to factor $12x^2-12x-72$ directly, without first pulling out 12, makes the numbers larger and increases the chance of a factoring error. Correct approach: always divide out any common constant first — giving $12(x^2-x-6)$ — then factor the simplified quadratic.

Mistake 2: Mixing up which test-interval a given sign combination belongs to. …