**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.

**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) = 2^{2} = 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 x_{0} ∈ (-∞, -1]

lim _{x-> x0} f(x) = lim _{x-> x0} 2x + 1

Applying the limit, we get

= 2x_{0} + 1 ------(1)

f(x_{0}) = 2x_{0} + 1 ------(2)

(1) = (2)

It is continuous in (-∞, -1].

let x_{0} ∈ (-1, -1)

lim _{x-> x0} f(x) = lim _{x-> x0} 3x

Applying the limit, we get

= 3x_{0} ------(1)

f(x_{0}) = 3x_{0} ------(2)

(1) = (2)

It is continuous in (-1, 1).

let x_{0} ∈ [1, ∞)

lim _{x-> x0} f(x) = lim _{x-> x0} 2x - 1

Applying the limit, we get

= 2x_{0 }- 1 ------(1)

f(x_{0}) = 2x_{0 }- 1 ------(2)

(1) = (2)

It is continuous in [1, ∞).

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

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

**Graph of f(x) = 3x :**

-1 < x < 1

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

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

x > = 1

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