How do you solve $2 X^{3} - x^{4} \leq 0$ $2X^3-x^4<=0$ ?

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How do you solve $2 X^{3} - x^{4} \leq 0$ $2X^3-x^4<=0$ ?

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Answer 1 · 2017-01-06T16:23:35

The values for $x$ $x$ that are solutions are $x \geq 2 and x \leq 0$ $x>=2" and "x<=0$

Written another way $x \in (- \infty, 0 .] and x \in [. 2, + \infty)$ $x in(-oo,0color(white)(.)] " and "x in [color(white)(.)2,+oo)$

Explanation:

Given: $2 x^{3} - x^{4} \leq 0$ $" "2x^3-x^4<=0$

$x^{3} (2 - x) \leq 0$ $x^3(2-x)<=0$

Suppose $x \geq 2$ $x>=2$ then $(2 - x) \leq 0$ $(2-x)<=0$ so $x^{3} (2 - x) \leq 0$ $x^3(2-x)<=0$
Consequently $x \geq 2$ $x>=2$ is part of the answer

Suppose $x = 0$ $x=0$ then $x^{3} (2 - x) = 0$ $x^3(2-x)=0$ which satisfies the $\leq 0$ $<=0$ criterion. Consequently $x = 0$ $x=0$ is part of the solution.

Suppose $x < 0$ $x<0$ then $x^{3} < 0$ $x^3<0$ and $(2 - x) > 0$ $(2-x)>0$ so their product is $< 0$ $<0$ thus satisfying the criterion $\leq 0$ $<=0$

So the value for $x$ $x$ that are solutions are $x \geq 2 and x \leq 0$ $x>=2" and "x<=0$

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