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Differentiability and continuity : If the function is continuous at a particular point then it is differentiable at any point at x=c in its domain. But the vice-versa is not always true. Differentiability implies continuity.1) The function 'f' is continuous at x = c that means there is no break in the graph at x = c.

2) In an interval, the function is said to be continuous if there is no gap in the graph of the function in the entire given interval.

3) The function 'f' is continuous, if it is continuous at every point in an open interval (a,b).

4) The function 'f' is continuous if,

$\lim_{x \rightarrow c^{-}}f(x)=\lim_{x \rightarrow c^{+}}f(x)=f(c)$

1) The function 'f' is discontinuous that means there is a break in the graph.

2) In an interval, the function is said to be discontinuous if there is gap in the graph of the function in the entire given interval.

3) The function 'f' is discontinuous, if it is discontinuous at every point in an open interval (a,b).

4) The function 'f' is discontinuous if,

$\lim_{x \rightarrow c^{-}}f(x)$ and $\lim_{x \rightarrow c^{+}}f(x)$ exists but are not equal.

5) $\lim_{x \rightarrow c^{-}}f(x)$ and $\lim_{x \rightarrow c^{+}}f(x)$ are exists and equal but are not equal to f(c).

6) f(c) is not defined.

LHL = $\lim_{x \rightarrow 2^{-}}\frac{f(x) -f(2)}{x -2}$

= $\lim_{x \rightarrow 2^{-}}\frac{|x -2| -0}{x -2}$

=-1 (derivative from left side)

And

RHL = $\lim_{x \rightarrow 2^{+}}\frac{f(x) -f(2)}{x -2}$

= $\lim_{x \rightarrow 2^{+}}\frac{|x -2| -0}{x -2}$

= 1 (derivative from right side)

Now derivative from left side and right side are not equal so the given function is not differentiable at x = 2 but from the graph we can see that the function is continuous at x = 2.

2) Determine the whether function is differentiable at x =2.

$ f(x)=\begin{bmatrix}x^{2}+1, & x\leq2 \\4x-3, & x>2 \end{bmatrix}$

First let us check whether the limit exists at x = 2

$\lim_{x \rightarrow c^{-}}f(x)=\lim_{x \rightarrow c^{+}}f(x) $

$\lim_{x \rightarrow c^{-}}f(x) = \lim_{x \rightarrow c^{-}}x^{2}+1 $

= $2^{2}$ + 1 =5

$\lim_{x \rightarrow c^{+}}f(x) = \lim_{x \rightarrow c^{-}}4x - 3 $

= 4(2) - 3 =5

So Left hand limit = Right hand limit $\Rightarrow$ = f(2)

So, limit exists at x= 2.

Now we will check the differentiability at x = 2

$\lim_{x \rightarrow 2^{-}}\frac{f(x)-f(2^{-})}{x-2}$

=$\lim_{x \rightarrow 2^{-}}\frac{x^{2}+1 -5}{x-2}$

=$\lim_{x \rightarrow 2^{-}}\frac{x^{2}-4}{x-2}$

=$\lim_{x \rightarrow 2^{-}}\frac{(x+2)(x-2)}{x-2}$

= $\lim_{x \rightarrow 2^{-}}(x+2)$

=4

$\lim_{x \rightarrow 2^{+}}\frac{f(x)-f(2^{+})}{x-2}$

=$\lim_{x \rightarrow 2^{+}}\frac{4x -3 -5}{x-2}$

=$\lim_{x \rightarrow 2^{+}}\frac{4x -8}{x-2}$

=$\lim_{x \rightarrow 2^{+}}\frac{4(x-2)}{x-2}$

= $\lim_{x \rightarrow 2^{+}}4$

=4

$\lim_{x \rightarrow 2^{-}}\frac{f(x)-f(2^{-})}{x-2}= \lim_{x \rightarrow 2^{+}}\frac{f(x)-f(2^{-})}{x-2}$ = 4

So the given function is differentiable at x= 2.

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