SKETCH THE GRAPH AND VERIFY 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)  =  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)  =  2x₀ - 1    ------(2)

(1)  =  (2)

It is continuous in [1, ∞).

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

Graph of f(x) = 3x :

-1 < x < 1 

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

x > = 1

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