Question

How do solve `x^2<=8x` and write the answer as a inequality and interval notation?

Answer 1

The answer is `x in [0 ,8 ]` or `0<=x<=8`

Explanation:

Let's rewrite the equation as

`x^2-8x<=0`, `=>`, `x(x-8)<=0`

Let `f(x)=x(x-8)`

Let's do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `0` `color(white)(aaaa)` `8` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-8` `color(white)(aaaaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

Therefore,

`f(x)<=0` when `x in [0 ,8 ]`, or `0<=x<=8`