# Thread: What is the domain of this function??

1. ## What is the domain of this function??

The domain of the function f(x)=

1-(2,+infinity)

2-all reals

3-(-infinity,-2)U(2,+infinity)

4-(-2,2)

5-(-infinity,-2)U(-2,+infinity)

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I try to solve this question :

4+x2>0

x2>-4
x>√-4 The square root is not defined for negative numbers.

2. Is it possible for $\displaystyle 4+x^2<4$?

3. Originally Posted by Plato
Is it possible for $\displaystyle 4+x^2<4$?

I don't understand your question..Can you clarify your idea??

4. Are the any values of $\displaystyle x$ which gives $\displaystyle 4+x^2 \le 0$?
If not, then the domain is all real numbers.

5. Originally Posted by Plato
Are the any values of $\displaystyle x$ which gives $\displaystyle 4+x^2 \le 0$?
If not, then the domain is all real numbers.
we don't have any value of x$\displaystyle 4+x^2 \le 0$

because the square root for negative number is not defined

6. Is this true, for all $\displaystyle x$ , $\displaystyle x^2 \ge 0$?
If that is true add 4 to both sides, $\displaystyle 4+x^2 \ge 4$.
Therefore, $\displaystyle \sqrt{4+x^2}$ is defined for all $\displaystyle x$.
Therefore, the domain of the function is all $\displaystyle x$.

7. Originally Posted by Plato
Is this true, for all $\displaystyle x$ , $\displaystyle x^2 \ge 0$?
If that is true add 4 to both sides, $\displaystyle 4+x^2 \ge 4$.
Therefore, $\displaystyle \sqrt{4+x^2}$ is defined for all $\displaystyle x$.
Therefore, the domain of the function is all $\displaystyle x$.

Thank you very much I understand your idea..

If I want to obtain the range of the following function

What I must do??

are there any steps to find range of any function ?and also the domain
f(x)=$\displaystyle \sqrt{1-x}+x^2$

8. Originally Posted by Diligent_Learner
Thank you very much I understand your idea..

If I want to obtain the range of the following function

What I must do??

are there any steps to find range of any function ?and also the domain
f(x)=$\displaystyle \sqrt{1-x}+x^2$
To find the domain,

$\displaystyle 1-x\ge 0$

$\displaystyle -x \ge -1$

$\displaystyle \boxed{x \le 1}$

To find the range,

$\displaystyle f(1)=\sqrt{1-1}+1^2=1$

$\displaystyle f(0)=\sqrt{1-0}+0^2=1$

$\displaystyle f(-1)=\sqrt{1-^-1}+(-1)^2=\sqrt{2}+1$

Therefore, $\displaystyle \boxed{f(x) \ge 1}$

9. Originally Posted by masters
To find the domain,

$\displaystyle 1-x\ge 0$

$\displaystyle -x \ge -1$

$\displaystyle \boxed{x \le 1}$

To find the range,

$\displaystyle f(1)=\sqrt{1-1}+1^2=1$

$\displaystyle f(0)=\sqrt{1-0}+0^2=1$

$\displaystyle f(-1)=\sqrt{1-^-1}+(-1)^2=\sqrt{2}+1$

Therefore, $\displaystyle \boxed{f(x) \ge 1}$
Thank you very much ..

I understand from your solution that:

If i want to find the range :

I should substitute the values of x that included into the domain into function

and then I notice the the y values that the function will take..