How do you find the domain and range for #(2/3)^x – 9#?

Question

Genesis Allen · Accepted Answer

#f(x) = (2/3)^x - 9#

#f(x)# is well defined for all #x in RR# so the domain is #RR#

By looking at end behaviour we find the range of #f(x)# is #(-9, oo)# As #x->-oo# we have #(2/3)^x = (3/2)^(-x) -> oo#, so #f(x) -> oo# As #x->oo# we have #(2/3)^x->0#, so #f(x) -> -9# So range #f(x) = (-9, oo)# plot{(2/3)^x - 9 [-22.5, 22.5, -11.25, 11.25]}

Abigail Allen · Answer

undefined To find the domain and range of the function $ f(x) = \left(\frac{2}{3}\right)^x - 9 $:

1. Domain:
The domain consists of all real numbers for which the function is defined. Since the base of the exponentiation is positive ($\frac{2}{3}$), the function is defined for all real numbers. Therefore, the domain is all real numbers, or $\mathbb{R}$.

2. Range:
To find the range, analyze the behavior of the function as $ x $ approaches positive and negative infinity. As $ x $ approaches positive infinity, $\left(\frac{2}{3}\right)^x$ approaches 0 because the base ($\frac{2}{3}$) is less than 1. Subtracting 9 from a value that approaches 0 results in negative values infinitely close to -9. Similarly, as $ x $ approaches negative infinity, $\left(\frac{2}{3}\right)^x$ approaches infinity, and subtracting 9 from a large positive value results in values approaching negative infinity. Therefore, the range is all real numbers less than or equal to -9, or $(-\infty, -9]$.