Question:
Consider the function f(x) given by:
f(x) = (x^2 + 3x - 4)/(x - 1)
(a) Determine the value of f(1) and explain whether the function is continuous at x = 1.
(b) Find any points of discontinuity for the function f(x) and classify the type of discontinuity at each point.
Answer:
(a) To find the value of f(1), substitute x = 1 into the expression for f(x):
f(1) = (1^2 + 3(1) - 4)/(1 - 1)
Simplifying the expression, we have:
f(1) = (1 + 3 - 4)/(1 - 1)
f(1) = 0/0
The result 0/0 is an undefined quantity, which indicates that the function is discontinuous at x = 1. However, it does not provide enough information to determine the type of discontinuity at this point.
(b) To find the points of discontinuity, we need to identify any values of x that make the denominator of the function equal to zero. In this case, the denominator is (x - 1). Set this equal to zero and solve for x:
x - 1 = 0
x = 1
So, x = 1 is a potential point of discontinuity.
Now, let's find the limit of f(x) as x approaches 1 from both the left and right sides to determine the type of discontinuity:
Left-hand limit:
lim x->1- (x^2 + 3x - 4)/(x - 1)
Substituting x = 1 - h (where h approaches 0), we have:
lim h->0- ((1 - h)^2 + 3(1 - h) - 4)/((1 - h) - 1)
Expanding and simplifying the expression:
lim h->0- (1 - 2h + h^2 + 3 - 3h - 4)/(1 - h - 1)
lim h->0- (-2h + h^2 - 4 + 3 - 3h)/(-h)
lim h->0- (-h^2 - 5h - 1)/(-h)
Now, we can simplify further by factoring out a negative sign from the numerator:
lim h->0- (-1)(h^2 + 5h + 1)/(-h)
Notice that when we cancel out the negatives, we are left with:
lim h->0- (h^2 + 5h + 1)/h
If we evaluate this limit using direct substitution, we get:
lim h->0- (0^2 + 5(0) + 1)/0 = 1/0 = ∞
The left-hand limit of f(x) as x approaches 1 is equal to infinity.
Right-hand limit:
lim x->1+ (x^2 + 3x - 4)/(x - 1)
Substituting x = 1 + h (where h approaches 0), we have:
lim h->0+ ((1 + h)^2 + 3(1 + h) - 4)/((1 + h) - 1)
Expanding and simplifying the expression:
lim h->0+ (1 + 2h + h^2 + 3 + 3h - 4)/(1 + h - 1)
lim h->0+ (h^2 + 5h)/h
Canceling out the h term, we obtain:
lim h->0+ (h + 5)
Evaluating this limit using direct substitution:
lim h->0+ (0 + 5) = 5
The right-hand limit of f(x) as x approaches 1 is equal to 5.
Since the left-hand limit (as x approaches 1) is infinity and the right-hand limit is 5, the two limits are not equal. Therefore, f(x) has a removable discontinuity at x = 1.
To summarize: