# SKETCH THE GRAPH AND VERIFY THE CONTINUITY OF THE FUNCTION

## About "Sketch the Graph and Verify the Continuity of the Function"

Sketch the Graph and Verify the Continuity of the Function :

Here we are going to see some example problems to understand the concept of sketching the graph and verifying the continuity of the function.

## Sketch the Graph and Verify the Continuity of the Function - Practice questions

Question 1 :

Let f(x)

Graph the function. Show that f(x) continuous on (- ∞, ∞).

Solution :

There are three partitions in the piecewise function.

To check the continuity at the point 0, we should prove the following.

lim x-> 0- f(x)  =  lim x->0+ f(x)  =  lim x->0 f(0)

lim x-> 0- f(x)  =  0   ----(1)

lim x->0+ f(x)  =  0   ----(2)

f(0)  =  0   ----(3)

(1)  =  (2)  =  (3)

Hence the function is continuous at x = 0.

Now let us check the continuity at the point 2.

lim x-> 2- f(x)  =  lim x->2+ f(x)  =  lim x->2 f(0)

lim x-> 2- f(x)  =  22  =  4   ----(1)

lim x->2+ f(x)  =  4   ----(2)

f(2)  =  4   ----(3)

(1)  =  (2)  =  (3)

Hence the function is continuous at x = 2.

So the given piecewise function is continuous on (- ∞, ∞).

Question 2 :

If f and g are continuous functions with f(3) = 5 and lim x->3 [2 f(x) - g(x)]  =  4, find g(3).

Solution :

lim x->3 [2 f(x) - g(x)]  =  4

Let us apply 3 in the function instead of x.

[2 f(3) - g(3)]  =  4

2(5) - g(3)  =  4

10 - g(3)  =  4

g(3)  =  10 - 4

g(3)  =  6

Hence the value of g(3) is 6.

Question 3 :

Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.

Solution :

First let us check the continuity at the point x  =  -1

lim x-> -1- f(x)  =  lim x-> -1- 2x + 1

By applying the limit, we get

=  2(-1) + 1

=  -2 + 1

=  -1 -----(1)

lim x-> -1+ f(x)  =  lim x-> -1+ 3x

By applying the limit, we get

=  3(-1)

=  -3 -----(2)

lim x-> -1- f(x)    lim x-> -1+

So, the function is not continuous at x = -1.

Now let us check the continuity at the point x  =  1

lim x-> 1- f(x)  =  lim x-> 1- 3x

By applying the limit, we get

=  3(1)

=  3 -----(1)

lim x-> -1+ f(x)  =  lim x-> -1+ 2x - 1

By applying the limit, we get

=  2(1) - 1

=  1 -----(2)

lim x-> 1- f(x)    lim x-> 1+

So, the function is not continuous at x = 1.

To find at which of these points f is continuous from the right, from the left, or neither, we have to draw the number line.

let x0 ∈ (-∞, -1]

lim x-> x0 f(x)  =  lim x-> x0 2x + 1

Applying the limit, we get

=  2x0 + 1  ------(1)

f(x0)  =  2x0 + 1  ------(2)

(1)  =  (2)

It is continuous in  (-∞, -1].

let x0 ∈ (-1, -1)

lim x-> x0 f(x)  =  lim x-> x0 3x

Applying the limit, we get

=  3x0  ------(1)

f(x0)  =  3x0  ------(2)

(1)  =  (2)

It is continuous in  (-1, 1).

let x0 ∈ [1)

lim x-> x0 f(x)  =  lim x-> x0 2x - 1

Applying the limit, we get

=  2x0 - 1  ------(1)

f(x0)  =  2x- 1    ------(2)

(1)  =  (2)

It is continuous in [1).

Graph of f(x) = 2x + 1 :

 x = -1f(-1)  =  -1 x = -2f(-2)  =  -3 x = -3f(-3)  =  -5

Graph of f(x) = 3x :

-1 < x < 1

 x = -0.5f(-0.5)  =  -1.5 x = -0.7f(-0.7)  =  -2.1 x = 0.5f(0.5)  =  1.5

Graph of f(x) = 2x - 1:

x > = 1

 x = 1f(1)  =  1 x = 2f(2)  =  3 x = 3f(3)  =  5

After having gone through the stuff given above, we hope that the students would have understood, "Sketch the Graph and Verify the Continuity of the Function"

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