Can you do the demonstration?

${ x }^{ 4 }-x\le 0$

${ x(x }^{ 3 }-1)\le 0$

So we have two ecuations ${ x }^{ 3 }-1\le 0$ and $x\le 0$

• If $x\le 0$ is true then ${ x(x }^{ 3 }-1)\le 0$ is fake because if $x<0$ then ${ x }^{ 3 }-1$ it wil be a negative number, and when we multiply two negative numbers, the result is a positive number and we need a negative number

• If ${ x }^{ 3 }-1\le 0$ is true we have two options, when ${ x }^{ 3 }-1=0$ and when ${ x }^{ 3 }-1<0$ , if ${ x }^{ 3 }-1=0$ is true, then $x=1$ so te original ecuation it will be true. And if ${ x }^{ 3 }-1<0$ is true then this factor become negative, and when we have ${ x(x }^{ 3 }-1)\le 0$ if ${ x }^{ 3 }-1$ is negative then when we multiply $x$ and ${ x }^{ 3 }-1$ the result it will be a negative number, so $-x\le 0$ and $x\ge 0$

So ${ x }^{ 3 }-1\le 0$ is true, then ${ x }^{ 3 }\le 1$ and ${ x }\le 1$ and we have $x\ge 0$ so ${ 0\le x }\le 1$

And the interval is [0,1] then the answer is 0.1