Answer:
False
Step-by-step explanation:
This is false. We can disprove this by plugging in the number into the function f(x) = lnx
f(0) = ln0
Since [tex]e^{- \infty} = 0[/tex], [tex]ln0 = -\infty[/tex]
This is not 0 so f(0) cannot be equal to 0, thus disproving the statement.
