For a function to be continuous at a point,
If the function f(x) is continuous at a point x = a, then
Problem 1 :
continuous at x = -1 and continuous at x = 2 ?
Solution :
If the function is continuous at x = -1, then
lim_{ x -> -1}^{-} f(x) = lim_{ x -> -1}^{+} f(x)
So, the function is continuous at x = -1.
If the function is continuous at x = 2, then
lim_{ x -> 2}^{-} f(x) = lim_{ x -> 2}^{+} f(x)
So, the function is not continuous at x = 2.
Problem 2 :
continuous at x = 1 ?
Solution :
So, the function g(x) is not continuous at x = 1.
Problem 3 :
Solution :
Checking if it is continuous at x = -3 :
Checking if it is continuous at x = 4 :
The function is not continuous at x = -3 and continuous at x = 4.
Problem 4 :
Solution :
Checking if it is continuous at x = π/2 :
Since the left hand limit and right hand limit they are not equal, f(x) is not continuous at x = π/2
Checking if it is continuous at x = π :
So, the function f(x) is not continuous at x=π/2 and continuous at x = π.
Problem 5 :
The function f has the properties indicated in the table below. Which of the following must be true ?
a) f is continuous at x = 1 b) f is continuous at x = 2
c) f is continuous at x = 3 d) None of the above.
Solution :
By observing the table, at x -> 3
lim_{ x->3}^{-} f(x) and lim_{ x->3}^{+} f(x) they are equal. So, the function f(x) is continuous at x --> 3
Problem 6 :
If the function f is continuous for all real numbers and if
f(x) = (x^{2} - 4)/(x + 2)
when x ≠ 2, then f(-2) =
A) -4 B) -2 C) -1 D) 0 E) 2
Solution :
f(x) = (x^{2} - 4)/(x + 2)
f(x) = (x^{2} - 2^{2})/(x + 2)
= (x + 2)(x - 2) / (x + 2)
f(x) = (x - 2)
f(-2) = -2 - 2
= -4
So, the answer is -4.
Problem 7 :
Let f be the function given above. What are all values of a and b for which f is differentiable at x = 2 ?
a) 1/4 and b = -1/2
b) a = 1/4 and b = 1/2
c) a = 1/4 and b is any real number]
d) a = b + 2 where b is any real number
e) There are no such values of a and b.
Solution :
lim _{x->2}- f(x) = lim _{x->2}- x + 2b
= 2 + 2b ----(1)
lim _{x->2}+ f(x) = lim _{x->2}+ ax^{2}
= a(2)^{2}
= 4a ----(2)
Since it is differentiable at x = 2
lim _{x->2} f(x) = f'(2)
f'(x) = 1 + 2(0)
f'(2) = 1 ----(3)
(1) = (3) 2 + 2b = 1 2b = -1 b = -1/2 |
(1) = (2) 2 + 2b = 4a 2 + 2(-1/2) = 4a 2-1 = 4a a = 1/4 |
So, option a is correct.
Dec 08, 23 08:03 AM
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