Q.Find intervals in which the function given by $f(x) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$ is

Concept: Monotonic Function Analysis — we find $f'(x)$, determine its sign intervals, and conclude where $f$ increases ($f'(x) > 0$) or decreases ($f'(x) < 0$).

Step 1: Differentiate

$f'(x) = \frac{3}{10} \cdot 4x^3 - \frac{4}{5} \cdot 3x^2 - 3 \cdot 2x + \frac{36}{5}$

$= \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5}$

Multiply through by 5:

$f'(x) = \frac{1}{5}(6x^3 - 12x^2 - 30x + 36)$

Step 2: Factor the cubic

$6x^3 - 12x^2 - 30x + 36 = 6(x^3 - 2x^2 - 5x + 6)$

Test $x=1$: $1 - 2 - 5 + 6 = 0$, so $(x-1)$ is a factor.

Divide: $x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6) = (x-1)(x-3)(x+2)$

Thus $f'(x) = \frac{6}{5}(x-1)(x-3)(x+2)$

Step 3: Sign analysis

Critical points: $x = -2, 1, 3$ (all real, simple roots).

The key idea is simple: a function is increasing where its derivative is positive, and decreasing where its derivative is negative. So the entire problem reduces to finding where $f'(x) > 0$ and where $f'(x) < 0$.

Why does this work? The derivative at a point tells you the slope of the tangent — if the slope is positive, the function is rising as you move right; if negative, it's falling. By finding where the derivative changes sign (at its zeros), we can break the real line into intervals and test each one.

Let's go step by step.

1. Compute the derivative carefully.

We have:

Differentiate term by term:

• $\frac{d}{dx}\left(\frac{3}{10}x^4\right) = \frac{3}{10} \cdot 4x^3 = \frac{12}{10}x^3 = \frac{6}{5}x^3$
• $\frac{d}{dx}\left(-\frac{4}{5}x^3\right) = -\frac{4}{5} \cdot 3x^2 = -\frac{12}{5}x^2$
• $\frac{d}{dx}(-3x^2) = -6x$
• $\frac{d}{dx}\left(\frac{36}{5}x\right) = \frac{36}{5}$
• $\frac{d}{dx}(11) = 0$

So:

2. Factor $f'(x)$ to find critical points.

Factor out $\frac{1}{5}$ to simplify:

Now factor the cubic inside. First, take out a common factor of 6:

So:

The sign of $f'(x)$ depends only on the cubic $p(x) = x^3 - 2x^2 - 5x + 6$, since $\frac{6}{5} > 0$.

3. Factor the cubic $p(x)$.

Try $x = 1$: $1 - 2 - 5 + 6 = 0$, so $x-1$ is a factor. Divide:

Now factor the quadratic:

Thus:

So the critical points (where $f'(x) = 0$) are $x = -2$, $x = 1$, and $x = 3$.

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

The critical points divide the real line into four intervals:

Pick a test point in each interval and evaluate the sign of $p(x) = (x-1)(x-3)(x+2)$. Remember, $f'(x) = \frac{6}{5}p(x)$, so the sign of $f'(x)$ is the same as the sign of $p(x)$.

Method: Testing Increasing/Decreasing Intervals via a Sign Chart

The general technique for locating every interval on which a function increases or decreases, using the sign of its derivative.

Steps

Step 1: Differentiate and factor completely

Compute $f'(x)$, then factor it as completely as possible into linear (or otherwise sign-determinable) factors. A fully factored derivative is what makes the sign chart mechanical instead of guesswork.

Step 2: Find every zero of $f'(x)$ — these are the boundary points

Solve $f'(x) = 0$ for all real roots. These critical points split the real line (or the given domain) into open intervals; the sign of $f'(x)$ can only change at one of these boundaries, never in the middle of an interval.

Step 3: Test the sign of each factor in every interval …

Common Mistakes

Mistake 1: Making a sign error when multiplying three negative factors

Why it's wrong: with $f'(x)=\frac{6}{5}(x-1)(x-3)(x+2)$, testing an interval like $(-\infty,-2)$ requires multiplying three negative quantities together; a common slip is stopping after two factors and concluding the product is positive, forgetting the third negative flips it back to negative. Correct approach: count the total number of negative factors in the product — an even count gives a positive product, an odd count gives a negative product.

Mistake 2: Including the critical points themselves in the increasing/decreasing intervals …