# How do you solve and write the following in interval notation: x^3 + 3x<4x^2 ?

Apr 12, 2017

Solution : $x < 0 \mathmr{and} 1 < x < 3$. In interval notation: $\left(- \infty , 0\right) \cup \left(1 , 3\right)$

#### Explanation:

${x}^{3} + 3 x < 4 {x}^{2} \mathmr{and} {x}^{3} + 3 x - 4 {x}^{2} < 0 \mathmr{and} x \left({x}^{2} - 4 x + 3\right) < 0$ or

$x \left(x - 1\right) \left(x - 3\right) < 0$. Critical points are $x = 0 , x = 1 , x = 3$

When $x < 0$ ::: $x \left(x - 1\right) \left(x - 3\right)$ is negative i.e $< 0$

When $0 < x < 1$ ::: $x \left(x - 1\right) \left(x - 3\right)$ is positive i.e $> 0$

When $1 < x < 3$ ::: $x \left(x - 1\right) \left(x - 3\right)$ is negative i.e $< 0$

When $x > 3$ ::: $x \left(x - 1\right) \left(x - 3\right)$ is positive i.e $> 0$

So Solution : $x < 0 \mathmr{and} 1 < x < 3$ in interval notation $\left(- \infty , 0\right) \cup \left(1 , 3\right)$ graph{x^3-4x^2+3x [-10, 10, -5, 5]}. The graph also confirms above result #[Ans]