Final Answer:
The graph of the function g(x) = (4x^2 + 3x + 1)/(x + 1) has a hole at x = -1. This means that there is a point at x = -1 where the function is undefined, but the limits of the function as x approaches -1 from both sides both exist and are equal. Option B is answer.
Step-by-step explanation:
For the function g(x) = (4x^2 + 3x + 1)/(x + 1) to be continuous at x = a, the following conditions must be met:
g(a) is defined.
lim_{x->a} g(x) exists.
lim_{x->a} g(x) = g(a).
Let's analyze each condition:
Condition 1: g(a) is defined
g(x) = (4x^2 + 3x + 1)/(x + 1) is defined for all real numbers except for x = -1, since the denominator cannot be zero. Therefore, g(-1) is undefined.
Condition 2: lim_{x->a} g(x) exists
To determine if lim_{x->a} g(x) exists, we can use direct substitution (if a ≠ -1) or l'Hôpital's rule (if a = -1).
Since a = -1 is undefined for g(x), we will use l'Hôpital's rule.
l'Hôpital's rule states that if lim_{x->a} f(x)/g(x) is of the form 0/0 or ∞/∞, then lim_{x->a} f(x)/g(x) = lim_{x->a} f'(x)/g'(x).
Applying l'Hôpital's rule to lim_{x->-1} g(x), we get:
lim_{x->-1} g(x) = lim_{x->-1} (4x^2 + 3x + 1)/(x + 1)
= lim_{x->-1} (8x + 3)/(1)
= 5
Condition 3: lim_{x->a} g(x) = g(a)
Since lim_{x->-1} g(x) = 5 and g(-1) is undefined, g(x) is not continuous at x = -1.
The function g(x) = (4x^2 + 3x + 1)/(x + 1) is not continuous at x = -1 due to a discontinuity of the type "hole." Option B is answer.