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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.

Last updated date: 20th Jun 2024
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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.