Question

Given the function f'(x) = x3 + 12x2 + 36x, determine all intervals on which f' is increasing

Answer 1

Answer:

$(-\infty,\, -6)$ and $(-2,\, \infty)$.

Step-by-step explanation:

A continuous function is increasing if and only if its first derivative is positive.

Differentiate $(x^{3} + 12\, x^{2} + 36\, x)$ with respect to $x$ to find the first derivative of this function.

$\begin{aligned}& \frac{d}{d x} \left[x^{3} + 12\, x^{2} + 36\, x\right] \\ =\; & 3\, x^{2} + 24\, x + 36 \\ =\; & 3\, (x^{2} + 8\, x + 12)\end{aligned}$.

In other words, the given function is increasing if and only if $3\, (x^{2} + 8\, x + 12) > 0$.

Notice that $12 = 2 \times 6$ while $(2 + 6) = 8$. Thus, the quadratic expression $3\, (x^{2} + 8\, x + 12)$ may be rewritten as:

$\begin{aligned} & 3\, (x^{2} + 8\, x + 12) \\ =\; & 3\, (x + 2) \, (x + 6)\end{aligned}$.

Thus, the requirement that $3\, (x^{2} + 8\, x + 12) > 0$ is equivalent to $3\, (x + 2) \, (x + 6) > 0$, which is true if and only if:

• either $(x + 2) < 0$ and $(x + 6) < 0$, or
• $(x + 2) > 0$ and $(x + 6) > 0$.

Equivalently, $3\, (x + 2) \, (x + 6) > 0$ is true if and only if:

• $x < (-2)$ and $x < (-6)$, or
• $x > (-2)$ and $x > (-6)$.

The first requirement simplifies to $x < (-6)$ and corresponds to the interval $(-\infty, \, -6)$.

The second requirement simplifies to $x > (-2)$, which corresponds to the interval $(-2,\, \infty)$.

Thus, the given function is increasing on the interval $(-\infty, \, -6)$ and $(-2,\, \infty)$.