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\[f(x)=\sqrt[4]{x^2+3x}\]

Hint: Inside cannot be negative.

I don't understand how to start the problem. :/

Because g can be zero and it can be positive, but it cannot be negative.

Like you know whatever g is. In this case your g is x^2+3x

Um. I'm not good at math so I'm having trouble understanding what you are saying. sorry.

Can you do fourth root of a negative number?
Would that exist?

I don't think you can do that.

oh. so I just solve \[x^2+3x \ge0\]?

That is right.

thank you. :D

Np. Do you know how to solve it?

I think I do haha. Let me try it...

Um. Okay. I don't know how to. ):

yeah it would be x=0, -3... i think.

Sorry. The site was updating. ):
The first and last intervals?

Thank you! Your explanation was really helpful. (: