How to find values of x that are not in the domain?

1 answer:

8 0

So the domain is "all x". To find the domain, I'll ignore the "x + 2" in the numerator (since the numerator does not cause division by zero) and instead I'll look at the denominator. I'll set the denominator equal to zero, and solve. The x-values in the solution will be the x-values which would cause division by zero.

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Write the first five terms of the the sequence defined by the explicit formula an=72(1/3)^n-1

lisabon 2012 [21]

$a_n=72\left(\dfrac{1}{3}\right)^{n-1}\\\\\text{Put}\ n=1,\ n=2,\ n=3,\ n=4,\ n=5\ \text{to the equation}:\\\\n=1\to a_1=72\left(\dfrac{1}{3}\right)^{1-1}=72\left(\dfrac{1}{3}\right)^0=72(1)=72\\\\n=2\to a_2=72\left(\dfrac{1}{3}\right)^{2-1}=72\left(\dfrac{1}{3}\right)^1=72\left(\dfrac{1}{3}\right)=\dfrac{72}{3}=24\\\\n=3\to a_3=72\left(\dfrac{1}{3}\right)^{3-1}=72\left(\dfrac{1}{3}\right)^2=72\left(\dfrac{1}{9}\right)=\dfrac{72}{9}=8\\\\n=4\to a_4=72\left(\dfrac{1}{3}\right)^{4-1}=72\left(\dfrac{1}{3}\right)^3=72\left(\dfrac{1}{27}\right)=\dfrac{72}{27}=\dfrac{8}{3}\\\\n=5\to a_5=72\left(\dfrac{1}{3}\right)^{5-1}=72\left(\dfrac{1}{3}\right)^4=72\left(\dfrac{1}{81}\right)=\dfrac{72}{81}=\dfrac{8}{9}\\\\Answer:\ \boxed{72,\ 24,\ 8,\ \dfrac{8}{3},\ \dfrac{8}{9}}$