Question

What is the sum of all real numbers x that are not in the domain of the function f(x) = \frac{1}{x^2-7} + \frac{1}{x^3-8} + \frac{1}{x^4-9}~?

Answer 1

Answer:

The sum of all real numbers x that are not in the domain of the function f(x) is 2.

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2-7}+\frac{1}{x^3-8}+\frac{1}{x^4-9}$

The domain is the set of inputs. So, the domain of the given function is all real numbers except those numbers for which denominator values is equal to 0.

$x^2-7=0$

$x^2=7$

$x=\pm \sqrt{7}$

It means $\pm \sqrt{7}$ is not included in domain.

$x^3-8=0$

$x^3=8$

$x=2$

It means 2 is not included in domain.

$x^4-9=0$

$x^4=9$

$x^2=\pm 3$

$x=\pm \sqrt{3}$

It means $\pm \sqrt{3}$ is not included in domain.

The sum of all real numbers x that are not in the domain of the function f(x) is

$Sum=-\sqrt{7}+\sqrt{7}+\sqrt{3}-\sqrt{3}+2=2$

Therefore the sum of all real numbers x that are not in the domain of the function f(x) is 2.