Q.Find the intervals in which the following functions are strictly increasing or decreasing:

Concept: Increasing Function Test

A function $f$ is strictly increasing where $f'(x) > 0$ and strictly decreasing where $f'(x) < 0$.

(a) $f(x) = x^2 + 2x - 5$

$f'(x) = 2x + 2 = 2(x+1)$. …

The core idea is simple: a function is strictly increasing where its derivative is positive, and strictly decreasing where its derivative is negative. This works because the derivative measures the instantaneous rate of change — if it's positive, the function is climbing; if negative, it's falling. We just need to find where the derivative changes sign, which happens at its zeros (critical points).

Let’s work through each part step by step.

(a) $f(x) = x^2 + 2x - 5$

1. Find the derivative:

   $f'(x) = 2x + 2 = 2(x + 1)$

2. Find critical points:

   Set $f'(x) = 0$: $2(x+1) = 0 \Rightarrow x = -1$. This is the only point where the derivative could change sign.

3. Test intervals around $x = -1$:

   • For $x < -1$, say $x = -2$: $f'(-2) = 2(-2+1) = -2 < 0$ → decreasing.
   • For $x > -1$, say $x = 0$: $f'(0) = 2(0+1) = 2 > 0$ → increasing.

4. Conclusion:

   Strictly decreasing on $(-\infty, -1)$, strictly increasing on $(-1, \infty)$.

(b) $f(x) = 10 - 6x - 2x^2$

1. Derivative:

   $f'(x) = -6 - 4x = -2(3 + 2x)$

2. Critical point:

   $-2(3 + 2x) = 0 \Rightarrow x = -\frac{3}{2}$

3. Test intervals:

   • For $x < -\frac{3}{2}$, say $x = -2$: $f'(-2) = -6 - 4(-2) = -6 + 8 = 2 > 0$ → increasing.
   • For $x > -\frac{3}{2}$, say $x = 0$: $f'(0) = -6 < 0$ → decreasing.

4. Conclusion:

   Increasing on $(-\infty, -\frac{3}{2})$, decreasing on $(-\frac{3}{2}, \infty)$.

(c) $f(x) = -2x^3 - 9x^2 - 12x + 1$

1. Derivative:

   $f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x+1)(x+2)$

2. Critical points:

   $-6(x+1)(x+2) = 0 \Rightarrow x = -1$ and $x = -2$.

3. Test intervals: The critical points divide the real line into three intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, \infty)$.

   • For $x < -2$, say $x = -3$: $f'(-3) = -6(-3+1)(-3+2) = -6(-2)(-1) = -12 < 0$ → decreasing.
   • For $-2 < x < -1$, say $x = -1.5$: $f'(-1.5) = -6(-1.5+1)(-1.5+2) = -6(-0.5)(0.5) = 1.5 > 0$ → increasing.
   • For $x > -1$, say $x = 0$: $f'(0) = -6(1)(2) = -12 < 0$ → decreasing.

4. Conclusion:

   Decreasing on $(-\infty, -2)$ and $(-1, \infty)$, increasing on $(-2, -1)$.

(d) $f(x) = 6 - 9x - x^2$

1. Derivative:

   $f'(x) = -9 - 2x = -(2x + 9)$

2. Critical point:

   $-(2x + 9) = 0 \Rightarrow x = -\frac{9}{2}$

3. Test intervals:

   • For $x < -\frac{9}{2}$, say $x = -5$: $f'(-5) = -9 - 2(-5) = -9 + 10 = 1 > 0$ → increasing.
   • For $x > -\frac{9}{2}$, say $x = 0$: $f'(0) = -9 < 0$ → decreasing.

4. Conclusion:

   Increasing on $(-\infty, -\frac{9}{2})$, decreasing on $(-\frac{9}{2}, \infty)$.

(e) $f(x) = (x+1)^3 (x-3)^3$

1. Simplify first:

   $f(x) = [(x+1)(x-3)]^3 = (x^2 - 2x - 3)^3$

2. Derivative using chain rule:

   $f'(x) = 3(x^2 - 2x - 3)^2 \cdot (2x - 2) = 3(x^2 - 2x - 3)^2 \cdot 2(x - 1) = 6(x-1)(x^2 - 2x - 3)^2$ …

Method: Classifying Increasing/Decreasing Intervals for Any Differentiable Function, Including Repeated Factors

This question covers several different function shapes (upward and downward parabolas, cubics, and a product with repeated linear factors) with one unified method: differentiate, factor completely, and read the sign of each factor — paying special attention to factors that appear to an even power, since those never change sign.

Steps

Step 1: Differentiate each function

Apply the standard rules (power rule, product/chain rule as needed) to get $f'(x)$ for the given expression.

Step 2: Factor $f'(x)$ into simplest linear (or repeated linear) factors

Pull out common constants, then factor any quadratic or higher-degree part fully. Watch for a derivative that comes out as a product involving a squared or otherwise even-power factor, such as $(x-a)^2$ — this happens naturally when the chain rule is applied to something like $[g(x)]^3$.

Step 3: Identify which factors can change sign and which cannot

A factor raised to an odd power (like $(x-a)$ or $(x-a)^3$) changes sign as $x$ crosses $a$. A factor raised to an even power (like $(x-a)^2$) is always $\ge 0$ and never changes sign — it only touches zero momentarily at $x=a$ without flipping. Only the odd-power factors actually control where the overall sign of $f'(x)$ switches.

Step 4: Build the sign chart using only the sign-changing factors …

Common Mistakes

Mistake 1: Treating a squared factor like $(x+1)^2(x-3)^2$ as changing sign at its roots

Why it's wrong: In part (e), $f'(x)=6(x-1)(x+1)^2(x-3)^2$ has factors $(x+1)^2$ and $(x-3)^2$ raised to even powers — these are always $\ge 0$ and never actually flip from positive to negative as $x$ crosses $-1$ or $3$, even though $f'(x)=0$ at those points. Treating them like ordinary sign-changing factors (as with $(x-3)$ in part (c)) leads to a wrong sign chart with extra false intervals. Correct approach: only the odd-power factor $(x-1)$ controls where the overall sign of $f'(x)$ actually switches; the squared factors just make the derivative momentarily touch zero without a genuine sign change.

Mistake 2: Mixing up the sign rule for a downward-opening quadratic derivative, as in parts (b) and (d) …