Question
Examine the continuity of the function f(x) = 2x2 – 1 at x = 3.
f(x) = 2x2 - 1
$^{\ \ \text{Lt}}_{\text{x}\rightarrow{3}}\text{f(x)}=^{\ \ \text{Lt}}_{\text{x}\rightarrow{3}}(2\text{x}^3-1) = 2(3)^2 - 1$
= 2(9) - 1 = 18 - 1 = 17
Now f is defined at x = 3
and f(x) = 2(3)2 - 1 = 2(9) - 1 = 18 - 1 = 17
$\therefore\ \ \text{Lt}\ \ \ \ \text{f(x)} = \text{f(3)} = 17\\ \ \ \ \ \text{x}\rightarrow3$
$\therefore$ f is continous at x = 3.
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