Answer:
a) decreasing
b) concave up
c) [tex]\lim_{{x \to +\infty}} f(x) = 6 \left( \lim_{{x \to +\infty}} \left(\dfrac{1}{9}\right)^x \right) = 6(0) = \textbf{0}[/tex]
d) [tex]\lim_{{x \to -\infty}} f(x) = 6 \left( \lim_{{x \to -\infty}}\left(\dfrac{1}{9}\right)^x \right) = 6(\infty) = \textbf{∞}[/tex]
Step-by-step explanation:
Given:
Exponential function [tex] f(x) = 6 \left(\dfrac{1}{9}\right)^x [/tex]:
a. Decreasing:
Since the base of the exponent [tex]\left(\dfrac{1}{9}\right)[/tex] is less than 1, the function will shrink as [tex]x[/tex] increases, making it decreasing.
b. Concave Up:
The base of the exponent [tex]\left(\dfrac{1}{9}\right)[/tex] is between 0 and 1, which means [tex] \left(\dfrac{1}{9}\right)^x [/tex] is concave up. Multiplying by a positive constant (6) doesn't change the concavity, so the function remains concave up.
c. [tex] \lim_{{x \to +\infty}} f(x) [/tex] as [tex] x [/tex] approaches positive infinity:
As [tex] x [/tex] approaches positive infinity, the term [tex] \left(\dfrac{1}{9}\right)^x [/tex] approaches 0.
So, the limit is:
[tex]\lim_{{x \to +\infty}} f(x) = 6 \left( \lim_{{x \to +\infty}} \left(\dfrac{1}{9}\right)^x \right) = 6(0) = \textbf{0}[/tex]
d. [tex] \lim_{{x \to -\infty}} f(x) [/tex] as [tex] x [/tex] approaches negative infinity:
As [tex] x [/tex] approaches negative infinity, the term [tex] \left(\dfrac{1}{9}\right)^x [/tex] approaches infinity. However, since the function is multiplied by a positive constant (6), the limit approaches positive infinity:
[tex]\lim_{{x \to -\infty}} f(x) = 6 \left( \lim_{{x \to -\infty}}\left(\dfrac{1}{9}\right)^x \right) = 6(\infty) = \textbf{∞}[/tex]
