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Question:

Examine the continuity of the function $f(x) = 2x^2 - 1$ at $x=3$

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Answer
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Solution:

Hint: To check for continuity at a point, check if the function exists at that point and if the limit of the function as x approaches that point is equal to the value of the function at that point.


Step-by-step Solution:

For $x = 3$:

1. The value of the function at $x = 3$ is:

$f(3) = 2(3^2) - 1 = 2(9) - 1 = 17$

So, $f(3)$ is defined.


2. If $lim_{x\to 3} f(x)$ exists: 

$lim_{x\to 3} f(x) = lim_{x\to 3} (2x^2 - 1)$

$\Rightarrow 2(3)^2 - 1 = 18 - 1 = 17$

Therefore, $lim_{x\to 3} f(x)$ exists and is equal to 17.


3. If $lim_{x\to 3} f(x) = f(3)$: 

$lim_{x\to 3} f(x) = 17$ and $f(3) = 17$. 

Therefore, $lim_{x\to 3} f(x) = f(3)$. 


Since all three conditions are met, we can conclude that the function $f(x) = 2x^2 - 1$ is continuous at $x=3$.


Note: A polynomial function, such as $f(x) = 2x^2 - 1$, is always continuous everywhere in its domain. The specific point mentioned was just to illustrate the method of proving continuity using the definition and limits.