**How to find points of discontinuity for a piecewise function :**

Here we are going to how to find out the point of discontinuity for a piecewise function.

Let us look into some examples to understand the concept.

**Question 1 :**

Find the points of discontinuity of the function f, where

**Solution :**

For the values of x greater than 3, we have to select the function 4x + 5.

lim _{x->3}- f(x) = lim _{x->3}- 4x + 5

= 4(3) + 5

= 12 + 5

= 17 -------(1)

For the values of x lesser than 3, we have to select the function 4x - 5.

lim _{x->3}+ f(x) = lim _{x->3}+ 4x - 5

= 4(3) - 5

= 12 - 5

= 7 -------(2)

lim _{x->3}- f(x) ≠ lim _{x->3}+ f(x)

Hence the function is not continuous at x = 3.

**Question 2 :**

Find the points of discontinuity of the function f, where

**Solution :**

For the values of x greater than 2, we have to select the function x + 2.

lim _{x->2}- f(x) = lim _{x->2}- x + 2

= 2 + 2

= 4 -------(1)

For the values of x lesser than 2, we have to select the function x^{2}.

lim _{x->2}+ f(x) = lim _{x->2}+ x^{2}

= 2^{2}

= 4-------(2)

lim _{x->2}- f(x) = lim _{x->2}+ f(x)

The function is continuous at x = 2.

Hence the given piecewise function is continuous for all x ∈ R.

**Question 3 :**

Find the points of discontinuity of the function f, where

**Solution :**

For the values of x greater than 2, we have to select the function x^{2} + 1

lim _{x->2}- f(x) = lim _{x->2}- x^{2} + 1

= 2^{2} + 1

= 5 -------(1)

For the values of x lesser than 2, we have to select the function x^{3 }- 3.

lim _{x->2}+ f(x) = lim _{x->2}+ x^{3 }- 3

= 2^{3 }- 3

= 8 - 3

= 5-------(2)

lim _{x->2}- f(x) = lim _{x->2}+ f(x)

The function is continuous at x = 2.

Hence the given piecewise function is continuous for all x ∈ R.

**Question 4 :**

Find the points of discontinuity of the function f, where

**Solution :**

Here we are going to check the continuity between 0 and π/2.

For the values of x lesser than or equal to π/4, we have to choose the function sin x.

lim _{x->}_{π/4-} f(x) = lim _{x->}π/4^{-} sin x

= sin (π/4)

= 1/√2

For the values of x greater than π/4, we have to choose the function cos x .

lim _{x->}_{π/4+} f(x) = lim _{x->}π/4^{+} cos x

= cos (π/4)

= 1/√2

The function is continuous for all x ∈ [0, π/2).

After having gone through the stuff given above, we hope that the students would have understood, "How To Find Points of Discontinuity For a Piecewise Function"

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