Find the domain of this function

[calif2a8]

What is the domain of sqrt(2x-x^3)

I thought this would be pretty straight forward but I am completely stumped. So obviously square root is not defined for (-inf, 0), therefore I said:

2x-x^3 >= 0

Working this out you get x <= sqrt(2) as the domain.

That is partially correct but I can't for the life of me get the rest of the answer.
The final answer is (-inf,-sqrt(2)] and [0, sqrt(2)]

HOW do you find out that (-sqrt(2), 0) is not defined?

Thanks for the read and help :D

 

[Mark44]

What is the domain of sqrt(2x-x^3)

I thought this would be pretty straight forward but I am completely stumped. So obviously square root is not defined for (-inf, 0), therefore I said:

2x-x^3 >= 0

Working this out you get x <= sqrt(2) as the domain.

You oversimplified things. I think this is what you did:
2x - x³ ≥ 0

x(2 - x²) ≥ 0 (so far, so good)

Next, it looks like you divided both sides by x - not good.

You need to factor the quadratic part, and you can't throw away the x factor as you did. Note that x = -2 is in the domain, which your work doesn't show.

That is partially correct but I can't for the life of me get the rest of the answer.
The final answer is (-inf,-sqrt(2)] and [0, sqrt(2)]

HOW do you find out that (-sqrt(2), 0) is not defined?

Thanks for the read and help :D

 

[calif2a8]

You oversimplified things. I think this is what you did:


x(2 - x²) ≥ 0 (so far, so good)

Actually I did something worse i think, but I didn't know it was not correct? I started with

2x - x³ ≥ 0

but then I did this: 2x ≥ x³ and then canceled the Xs...
so then I was left with

2 ≥ x² ... leading to my original answer of sqrt(2) ≥ x

Why is this approach wrong?

Thanks for all the replies by the way :)

 

[Mentallic]

Actually I did something worse i think, but I didn't know it was not correct? I started with

2x - x³ ≥ 0

but then I did this: 2x ≥ x³ and then canceled the Xs...
so then I was left with

2 ≥ x² ... leading to my original answer of sqrt(2) ≥ x

Why is this approach wrong?

Thanks for all the replies by the way :)

Dividing through by x before or after moving the x³ to the other side is an equivalent procedure.

What you should do is draw the graph of $y=x^3-2x$ and notice where y>0. You should know that $y=x(x^2-2)$ is the same graph. Now draw just $y=x^2-2$. Although the shape is completely different, the part where it crosses the x-axis at $x=\pm\sqrt{2}$ is still the same with both graphs, but it's missing some extra information, such as it doesn't cross the x-axis at x=0.

Why's this? Well, when you divided $x^3-2x$ by x, this assumes that $x\neq 0$ because you can't divide by zero, but x can be equal to zero, so you've changed the graph.

By the way, if $x^2\leq 2$ then the solution set for this inequality is [-2,2]. Again, draw the graph of $y=x^2-2$ and find where y<0.