Find the Value of b That Makes the Function Continuous :
Here we are going to see how to find the value of b that makes the function continuous.
Question 1 :
Find the constant b that makes g continuous on (−∞, ∞)
Solution :
Since the function is continuous at the point x = 4, then
lim _{x-> 4-} f(x) = lim _{x-> 4+} f(x)
lim _{x-> 4-} f(x) = lim _{x-> 4-} x^{2} - b^{2}
= 4^{2} - b^{2}
= 16 - b^{2} ----(1)
lim _{x-> 4+} f(x) = lim _{x-> 4+-} bx + 20
= b(4) + 20
= 4b + 20 ----(2)
(1) = (2)
16 - b^{2 } = 4b + 20
b^{2} + 4b + 20 - 16 = 0
b^{2} + 4b + 4 = 0
(b + 2)(b + 2) = 0
b + 2 = 0
b = -2
Hence the value of b is -2.
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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