# What is the domain and range of g(x) = 1/(7-x)^2?

Aug 13, 2015

Domain: $\left(- \infty , 7\right) \cup \left(7 , + \infty\right)$.
Range: $\left(0 , + \infty\right)$

#### Explanation:

The domain of the function will have to take into account the fact that the denominator cannot be equal to zero.

This means that any value of $x$ that will make the denominator equal to zero will be excluded from the domain.

${\left(7 - x\right)}^{2} = 0 \implies x = 7$

This means that the domain of the function will be $\mathbb{R} - \left\{7\right\}$, or $\left(- \infty , 7\right) \cup \left(7 , + \infty\right)$.

To find the range of the function, first note that a fractional expression can only be equal to zero if the numerator is equal to zero.

In your case, the numberator is constant and equal to $1$, which means that you cannot find an $x$ for which $g \left(x\right) = 0$.

Moreover, the denominator will always be positive, since you're dealing with a square. This means that the range of the function will be $\left(0 , + \infty\right)$.

graph{1/(7-x)^2 [-20.28, 20.27, -10.14, 10.12]}