Find the value of k that will make fx continuous everywhere

Find the value of k that will make f(x) continuous everywhere.

f(x) = { 2x-3    , x < or equal to 2

         { ln(x+k), x>2

This is supposed to be a piecewise function, but couldn\'t figure out the best way to type it. If someone could explain the steps to me, that would be great!

Solution

Answer : k = e-2 = 0.718

>> (2x-3) is continuous for all values of x <=2.

The only thing stopping this function being continuous everywhere is the point on the boundaries on the two function pieces. i.e we want, when x = 2,

2x-3 = ln(x+k)

=> 2*2-3 = ln(2+k)

=> 1 = ln(2+k)

=>2+k = e

=> k = e-2

=> k = 0.718

Hence k = 0.718 make f(x) continuous everywhere.