82,415 views
8 votes
8 votes
For which of the following values of x is the function f(x) = 14- X2 NOT defined as a real number? [square root] A. -2 B. 0 C. 2 D. 4

User Bryan Shalloway
by
2.7k points

1 Answer

9 votes
9 votes

We want to find the value for which the function,


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

not defined as a real number.

The square root of a number is a real number when the number is a positive number.

So, if we put in the values of x , we see that.


\begin{gathered} \sqrt[]{4-(-2)^2}=\sqrt[]{4-4}=0\text{ this is a real number } \\ \sqrt[]{4-(0)^2}=\sqrt[]{4}=2\text{ this is also a real number} \\ \sqrt[]{4-(2)^2}=\sqrt[]{4-4}=0\text{ this is a real number too} \\ \sqrt[]{4-(4)^2}=\sqrt[]{4-16}=\sqrt[]{-12}\text{ this is NOT a real number} \end{gathered}

We see that, the value of x that does not return a real number is x = 4, Option DW

User Laridzhang
by
2.8k points