Question

examine the continuity of the function f(x)=2x^(2)-1 at x=-2​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \bull \: f(x) = {2x}^{2} - 1$

$\large\underline{\sf{To\:discuss - }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \bull \sf \: \: The \: continuity \: of \: f(x) \: at \: x = - 2$

$\large\underline{\sf{Solution-}}$

Definition of Continuity :-

A function f(x) is said to be Continuous at x = a iff

$\sf \: f( a) = \:\lim_{x \: \to \: a^-}f(x) = \: \lim_{x \: \to \: a^ + }f(x)$

Now,

Given that

$\sf \: f(x) = {2x}^{2} - 1$

So,

$\rm :\longmapsto\:f( - 2) = 2 {( - 2)}^{2} - 1$

$\bf :\longmapsto\:f( - 2) = 8 - 1 = 7 - - (1)$

Now,

Left Hand Limit at x = - 2

$\rm :\longmapsto\:\sf \:\lim_{x \: \to \: - 2^-}f(x)$

$:\longmapsto\:\sf \: = \: \:\lim_{x \: \to \: - \: 2^-}( {2x}^{2} - 1)$

$\bf \: Put \: x \: = - 2 - h , As \: x \to \: - 2 \: so \: h \to \: 0$

$\longmapsto\:\sf \: \: = \: \lim_{h \: \to \: 0}\bigg(2 {( - 2 - h)}^{2} - 1 \bigg)$

$\rm :\longmapsto\: \: = \: 2 {( - 2)}^{2} - 1$

$\rm :\longmapsto\: = \: 8 - 1$

$\rm :\longmapsto\: = \: 7 \: - - (2)$

Now,

Right Hand Limit at x = - 2

$\longmapsto\:\sf \:\lim_{x \: \to \: 2^ + }f(x)$

$:\longmapsto\:\sf \: = \: \:\lim_{x \: \to \: - \: 2^ + }( {2x}^{2} - 1)$

$\bf \: Put \: x \: = - 2 + h , As \: x \to \: - 2 \: so \: h \to \: 0$

$\longmapsto\:\sf \: \: = \: \lim_{h \: \to \: 0}\bigg(2 {( - 2 + h)}^{2} - 1 \bigg)$

$\rm :\longmapsto\: \: = \: 2 {( - 2)}^{2} - 1$

$\rm :\longmapsto\: = \: 8 - 1$

$\rm :\longmapsto\: = \: 7 \: - - (3)$

From equation (1), (2) and (3), we concluded that

$\sf \: f( - 2) = \:\lim_{x \: \to \: 2^-}f(x) = \: \lim_{x \: \to \: 2^ + }f(x)$

$\bf\implies \:f(x) = {2x}^{2} - 1 \: is \: continuous \: at \: x = - 2$

Additional Information :-

A function f(x) is said to be continuous on a closed interval [a, b] if the following conditions are satisfied:

• f(x) is continuous from the right at a;

• f(x) is continuous from the left at b.

If the function f and g are continuous at c then

• f + g is continuous at c;

• f - g is continuous at c;

• f.g is continuous at c;

   

• f/g is continuous at c if g(c) ≠ 0 and is discontinuous at c if g(c) = 0