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Question:

Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

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Last updated date: 20th Jun 2024
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Answer
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Solution:

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.

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