Q.Find the intervals in which the function $f$ given by $f(x) = 2x^3 - 3x^2 - 36x + 7$ is

Concept: Increasing Function Test — $f'(x) > 0$ on an interval means $f$ is increasing there; $f'(x) < 0$ means decreasing.

Step 1: Differentiate

$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$.

Step 2: Find critical points

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

Step 3: Test sign of $f'(x)$ in each interval

• For $x < -2$ (e.g., $x=-3$): $(x-3)(x+2) = (-)(-) = +$, so $f'(x) > 0$ → increasing. …

The key idea is the Increasing Function Test: if $f'(x) > 0$ on an interval, $f$ is increasing there; if $f'(x) < 0$, $f$ is decreasing. This works because the derivative tells us the slope of the tangent — positive slope means the graph is rising as we move right, negative slope means it's falling.

For a polynomial like this, $f'$ is continuous, so it can only change sign at its zeros. Our job is to find where $f'$ is positive and where it's negative.

1. Find the derivative. Differentiate term by term:

Factor out the common 6:

2. Factor the quadratic. $x^2 - x - 6 = (x-3)(x+2)$. So

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

These are the only places where $f'$ could change sign.

4. Test the sign of $f'$ in each interval.

   The real line is split into three intervals by $-2$ and $3$: $(-\infty,-2)$, $(-2,3)$, and $(3,\infty)$. Pick a test point in each — choose numbers that are easy to evaluate.

   • Interval $(-\infty,-2)$: pick $x=-3$.

     $f'(-3) = 6(-3-3)(-3+2) = 6(-6)(-1) = 36 > 0$.

     So $f$ is increasing here.

   • Interval $(-2,3)$: pick $x=0$.

     $f'(0) = 6(0-3)(0+2) = 6(-3)(2) = -36 < 0$.

     So $f$ is decreasing here.

   • Interval $(3,\infty)$: pick $x=4$.

     $f'(4) = 6(4-3)(4+2) = 6(1)(6) = 36 > 0$.

     So $f$ is increasing here. …

Method: Finding Intervals of Increase/Decrease for a Cubic (Factor-and-Test Method)

A cubic function's derivative is a quadratic, which can have up to two critical points, splitting the domain into three intervals. Factoring the derivative is the key step that makes testing each interval fast and reliable.

Steps

Step 1: Differentiate the cubic

Compute $f'(x)$; it will be a quadratic expression in $x$.

Step 2: Factor $f'(x)$ completely

Pull out any common numeric factor first, then factor the remaining quadratic into two linear factors, e.g. $f'(x) = k(x - r_1)(x - r_2)$. Factoring makes the sign of $f'(x)$ at any point a simple product of signs, instead of a messy arithmetic evaluation.

Step 3: Locate the critical points and set up the three intervals

Setting $f'(x) = 0$ gives the two roots $r_1$ and $r_2$ (assume $r_1 < r_2$); these split the real line into $(-\infty, r_1)$, $(r_1, r_2)$, and $(r_2, \infty)$.

Step 4: Test the sign of each linear factor in every interval …

Common Mistakes

Mistake 1: Sign error when multiplying the two linear factors in the sign chart

Why it's wrong: With $f'(x)=6(x-3)(x+2)$, evaluating the sign in the middle interval $(-2,3)$ requires correctly tracking that $(x-3)$ is negative there while $(x+2)$ is positive — a slip in either sign flips the conclusion for that whole interval. Correct approach: substitute an actual test value (e.g. $x=0$) into each factor separately and multiply the signs explicitly, rather than reasoning about the signs abstractly.

Mistake 2: Forgetting that a cubic's derivative can have two critical points, not one …