Express the domain of f(x)= square root of (4-3x-x^2)?
\[f(x)=\sqrt{4-3x-x^2}\] \[=\sqrt{(4+x )(1-x)}\] What can't x be? You know that the number inside the square root cannot be negative.
What is the smallest x-value that DOES NOT make the number inside the square root sign a negative?
You have the number 0 inside the square root sign but not less than that or else it will be a negative.
I need to express the domain of f(x) =\[\sqrt{4-3x-x^2}\] in interval notation? It's for calculus
We're taking it step by step. You're after this right? \[x>?\]
We're taking it step by step here mate.
What x value makes the whole term 0?
@Jilian
1 makes the whole term zero
Good so x has to be greater than or equal to 1.
\[x \ge1\]
Do you understand these concepts of intervals now?
If you go below 1 you will find yourself having a value less than 0 inside the square root which you don't want.
Anything greater than one will make the solution non existing. I'm not clear on this interval notation however.
less than one sorry
My bad. I thought it was x-1 rather than 1-x
Apologies.
\[x \le1\]
Now for the 4+x
What's the maximum negative number you can put in their that won't make it negative. SO what number in there would make the value become zero?
-4 makes ( 4+x) = 0
Good work. so then x can't be less than -4. So then, x has to be greater than or equal to -4. \[x\ge -4\]
Now you have these two inequalties. \[x\le 1\] \[x\ge -4\] Is x ranging between these two inequalities?
so the domain will look like this, \[-4\le x \le1\]
You owned it. Well Done. Excellent work mate.
Thank You very much for your help. It was a pleasure working with you. You should be a Math professor.
Now worries mate, and hahah, nah, I don't think so.
No worries*
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