For each line, state its slope and its y-intercept. x+4y=1
1 answer:
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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>
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
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>
