consider the function f(x) = aln(x+2). Given that f'(1) = a/3, what is the approximate value of f(0.98)?
f(x1) = f(x0) + f'(x0)x(x1-x0)
The Attempt at a Solution
I solved it and get
f(.98) = aln(1+2) + (.098-1) = aln(3) - (.02)(a/3) <= not an answer
the bad thing is my teacher don't give me the increment interval. ..So I assumed it is .02