For some function L(x), the function p(x) is defined by the formula
p(x)={ x^2+3 for (-inf,-1), L(x) for [-1,3], x*2^x for (3,inf)}
Find the expression for the linear function L(x) that makes p(x) continuous.
Any help or hints on this problem will be greatly appreciated :)
p(x) = x2 + 3 for -infinity < x < -1
L(x) = ? for -1 <= x <= 3
p(x) = x·2x for 3 < x < infinity
The left-hand endpoint for L(x) has an x-value of -1.
Therefore, it must have the same value as what p(x) would have if the domain of p(x) included -1:
p(-1) = (-1)2 + 3 = 1 + 3 = 4 ---> the point is (-1, 4)
The right-hand endpoint for L(x) has an x-value of 3.
Therefore, it must have the same value as what p(x) would have if the domain of p(x) included 3:
p(3) = 3·23 = 3·8 = 24 ---> the point is (3, 24).
For L(x), you'll need to find the linear equation that includes the points (-1, 4) and (3, 24).
Can you finish this?