+0

# PLZ HALP!!

0
72
1

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 :)

Jun 11, 2020

#1
0

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?

Jun 11, 2020