Question

Where is the function 3(x + 4)(x – 7)^3 > 0?

Answer 1

Answer::

`(-\infty,-4)\cup(7,\infty)`

Explanation:

Given the function:

`3\left(x+4\right)\left(x-7\right)^3`

We want to determine the interval in which the function is greater than 0.

First, solve for the critical values.

`\begin{gathered} 3\left(x+4\right)\left(x-7\right)^3=0 \\ x+4=0,x-7=0 \\ x=-4,x=7 \end{gathered}`

So, the following are the possible intervals:

`\begin{gathered} (-\infty,-4) \\ (-4,7) \\ (7,\infty) \end{gathered}`

We test each of the intervals:

`\begin{gathered} (-\infty,-4),\text{ At }x=-5:3\left(x+4\right)\left(x-7\right)^3=5184>0 \\ (-4,7),\text{ At x}=0:3\left(x+4\right)\left(x-7\right)^3=-4116<0 \\ (7,\infty),\text{ At }x=8:3\left(x+4\right)\left(x-7\right)^3=36>0 \end{gathered}`

Therefore, the function is greater than zero in the intervals:

`(-\infty,-4)\cup(7,\infty)`

The result can be confirmed using the graph below: