The function is
.
To the left of 1 the function is a quadratic polynomial, to the right, it is a linear polynomial. Polynomial functions are always continuous, so the only candidate point for discontinuity is x=1.
The left limit is calculated with 1 substituted in
, which gives 5.
The right limit, is computed using the rule for the right part of 1, that is x+4.
Thus, the right limit is 1+4=5.
So, both left and right limits are equal. Now if f(1) is 5, then the function is continuous at 1.
But the function is not defined for x=1, that is x=1 is not in the domain of the function. Thus, we have a "whole" (a discontinuity) in the graph of the function.
The reason is now clear:
Answer:<span> f(1) is not defined</span>
Answer:
y = 1/2x - 6
Step-by-step explanation:
y2 - y1 / x2 - x1
-3 - (-5) / 6 - 2
2/4
= 1/2
y = 1/2x + b
-3 = 1/2(6) + b
-3 = 3 + b
-6 = b
A,C,and D are all correct
3(2x + 4) = -6
6x + 12 = -6
6x = -6 - 12
6x = - 18
x = -18/6
x = -3