Question

Find the interval in which the function \[ f(x) = 4x^3 - 6x^2 - 72x + 30 \] is (i) increasing and (ii) decreasing.

Hint:

Use first derivative test: $f'(x) > 0$ → Increasing, $f'(x) < 0$ → Decreasing.

Answer 1

Step 1: Differentiate the function.
\[ f(x) = 4x^3 - 6x^2 - 72x + 30 \] \[ f'(x) = 12x^2 - 12x - 72 \]

Step 2: Simplify derivative.
\[ f'(x) = 12(x^2 - x - 6) = 12(x-3)(x+2) \]

Step 3: Critical points.
Set $f'(x) = 0$: \[ 12(x-3)(x+2) = 0 $\Rightarrow$ x = 3, \; x = -2 \]

Step 4: Sign analysis of $f'(x)$.
Check intervals $(-\infty,-2)$, $(-2,3)$, and $(3,\infty)$: \[\begin{array}{rl} \bullet & \text{For $x < -2$: Choose $x = -3$, then $(x-3)(x+2) = (-6)(-1) = +6 > 0$.} \\ \bullet & \text{For $-2 < x < 3$: Choose $x = 0$, then $(x-3)(x+2) = (-3)(2) = -6 < 0$.} \\ \bullet & \text{For $x > 3$: Choose $x = 4$, then $(x-3)(x+2) = (1)(6) = +6 > 0$.} \\ \end{array}\]

Step 5: Conclusion.
\[\begin{array}{rl} \bullet & \text{Function is increasing in intervals $(-\infty, -2)$ and $(3, \infty)$.} \\ \bullet & \text{Function is decreasing in interval $(-2, 3)$.} \\ \end{array}\]

Final Answer: \[ \boxed{\text{Increasing in } (-\infty, -2) \cup (3,\infty), \text{Decreasing in } (-2,3)} \]