Question

question 6 (4 points)for what interval(s) is the function $f(x) = x^3 - x^2 - 6x$ increasing?$(-\infty, 1.8)$$(-1, \infty)$$(-1, 1.8)$$(-\infty, -1.1)$ and $(1.8, \infty)$

Explanation:

Step1: Find first derivative

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

Step2: Solve $f'(x)=0$

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $3x^2-2x-6=0$, where $a=3$, $b=-2$, $c=-6$:

$$\begin{align*} x&=\frac{2\pm\sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\ &=\frac{2\pm\sqrt{4+72}}{6}\\ &=\frac{2\pm\sqrt{76}}{6}\\ &=\frac{2\pm2\sqrt{19}}{6}\\ &=\frac{1\pm\sqrt{19}}{3} \end{align*}$$

Calculate approximate values: $\frac{1+\sqrt{19}}{3}\approx1.8$, $\frac{1-\sqrt{19}}{3}\approx-1.1$

Step3: Test sign of $f'(x)$

• For $x<-1.1$, $f'(x)>0$ (function increasing)
• For $-1.1
• For $x>1.8$, $f'(x)>0$ (function increasing)

Answer:

$(-\infty, -1.1)$ and $(1.8, \infty)$