Question
Examine the continuity of the function $f(x) = 2x^2 - 1$ at $x = 3.$
✓
Answer
Given: $f\left( x \right) = 2{x^2} - 1$
Continuity at $x = 3 \mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - 1} \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
And $f\left( 3 \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
Since $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = f\left( x \right),$
therefore, $f(x)$ is continuous at $x = 3.$
Continuity at $x = 3 \mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - 1} \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
And $f\left( 3 \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
Since $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = f\left( x \right),$
therefore, $f(x)$ is continuous at $x = 3.$
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