How to Determine If a Function is Continuous on a Graph :
Here we are going to see how to determine if a function is continuous on a graph.
Question 1 :
State how continuity is destroyed at x = x_{0 }for each of the following graphs.
(i)
Solution :
By observing the given graph, we come to know that
lim_{ x-> x0- }f(x) = f(x_{0}) (Because we have filled circle)
But,
lim_{ x-> x0+ }f(x) ≠ f(x_{0}) (Because we have unfilled circle)
Hence the given function is not continuous at the point x = x_{0}.
(ii)
Solution :
By observing the given graph, we come to know that
lim_{ x-> x0- }f(x) = f(x_{0}) (Because we have unfilled circle)
But,
lim_{ x-> x0+ }f(x) = f(x_{0}) (Because we have the same unfilled circle at the same place)
Hence the given function is continuous at the point x = x_{0}.
(iii)
Solution :
From the given picture, we know that lim_{ x-> x0- }f(x) = -∞
But,
lim_{ x-> x0- }f(x) = -∞
Hence it is not continuous at x = x_{0.}
(iv)
Solution :
lim_{ x-> x0- }f(x) = f(x_{0}) (Because we have unfilled circle)
But,
lim_{ x-> x0+ }f(x) ≠ f(x_{0}) (Because we have filled circle at different place)
Hence the given function is not continuous at the point x = x_{0}.
Question 2 :
Consider the function f (x) = x sin π/x What value must we give f(0) in order to make the function continuous everywhere?
Solution :
f (x) = x sin π/x
Range of sin x is [-1, 1]
-1 ≤ sin π/x ≤ 1
By multiplying x throught the equation, we get
-x ≤ x (sin π/x) ≤ x
Now let us apply the limit values
lim _{x -> 0} (-x) ≤ lim _{x -> 0 }x (sin π/x) ≤ lim _{x -> 0 }x
0 ≤ lim _{x -> 0 }x (sin π/x) ≤ 0
By sandwich theorem
lim _{x -> 0 }x (sin π/x) = 0
Now let us redefine the function
From this we come to know the value of f(0) must be 0, in order to make the function continuous everywhere
Question 3 :
The function f(x) = (x^{2} - 1) / (x^{3} - 1) is not defined at x = 1. What value must we give f(1) inorder to make f(x) continuous at x = 1 ?
Solution :
By applying the limit value directly in the function, we get 0/0.
Now let us simplify f(x)
f(x) = (x^{2} - 1) / (x^{3} - 1)
= (x + 1) (x - 1)/(x - 1)(x^{2} + x + 1)
= (x + 1) / (x^{2} + x + 1)
lim _{x-> 1} f(x) = lim _{x-> 1} (x + 1) / (x^{2} + x + 1)
= (1 + 1)/ (1 + 1 + 1)
= 2/3
By redefining the function, we get
From this we come to know the value of f(1) must be 2/3, in order to make the function continuous everywhere
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