VerifiedSolution:
Hint: To prove that a function f(x) is continuous at a point x = a, we need to show that the following two conditions are met:
$lim_(x \rightarrow a) f(x)$ exists
$lim_(x \rightarrow a) f(x) = f(a)$
Step-by-step Solution:
For $ x = 0 $:
1. The value of the function at $ x = 0 $ is:
$ f(0) = 5(0) - 3 = -3 $
So, $ f(0) $ is defined.
2. The limit of $ f(x) \ as\ \ x $ approaches 0 is:
$ \lim_{{x \to 0}} (5x - 3) = 5(0) - 3 = -3 $
Since $ \lim_{{x \to 0}} f(x) = f(0) = -3 $, the function is continuous at $ x = 0 $.
For $ x = -3 $:
1. The value of the function at $ x = -3 $ is:
$ f(-3) = 5(-3) - 3 = -18 $
So, $ f(-3) $ is defined.
2. The limit of $ f(x) \ as\ \ x $ approaches -3 is:
$ \lim_{{x \to -3}} (5x - 3) = 5(-3) - 3 = -18 $
Since $ \lim_{{x \to -3}} f(x) = f(-3) = -18 $, the function is continuous at $ x = -3 $.
For $ x = 5 $:
1. The value of the function at $ x = 5 $ is:
$ f(5) = 5(5) - 3 = 22 $
So, $ f(5) $ is defined.
2. The limit of $ f(x)\ as\ \ x $ approaches 5 is:
$ \lim_{{x \to 5}} (5x - 3) = 5(5) - 3 = 22 $
Since $ \lim_{{x \to 5}} f(x) = f(5) = 22 $, the function is continuous at $ x = 5 $.
Note: The function f(x) = 5x – 3 is continuous for all real numbers. This is because it is a polynomial function, and polynomial functions are continuous for all real numbers.
