In a section of a calculus workbook dealing with local linearity and linear approximations of functions, the following question is posed:
Consider the function f(x) = aln(x+2). Given that f'(1) = a/3, what is the approximate value of f(0.98)?
(A) (a/3)*(-0.02) + aln(2.98)
(B) -0.02a/3
(C) (0.98)*ln(a/3)
(D) (a/3)aln(2.98) + 0.98
(E) (a/3)*0.98 + aln(2.98)
When I look for an equation of a line tangent to f(x) at x=1, I get y=(a/3)(x-1) + aln3. Plugging in x=0.98, I get (a/3)(-0.02) + aln3, which seems to correspond to none of the answer choices . . .
(A) seems the closest, and therefore would give an "approximate" value, but if they're using the slope (a/3) why are they not then using the y-value at that the point (1, aln3) with that slope, which is f(1) = aln(1+2) = 3?
If we say x = 1 and x' = 0.98, then the formula they seem to be using to approximate f(x') is f(x')≈f'(x)(x'-x) + f(x') which seems sort of silly, because it assumes we already know what f(x'), in which case, why would we be looking for it? The way I learned to do it is f(x')≈f'(x)(x'-x) + f(x).
I'll redo everything, just to make sure that you are correct.
If $f(x) =a\ln(x+2) $, $f'(x) =\frac{a}{x+2} $. So, $f'(1) =\frac{a}{3} $. This is therefore a consequence of the definition of $f(x)$, so I do not see why it is "given".
Anyway, for a nicely behaved function (which this one is), $f(x+h) \approx f(x)+hf'(x) $. Putting $x=1$ and $h=-.02$, $f(0.98) \approx f(1)-0.02 f'(1) = a\ln(3)-0.02\frac{a}{3} = a(\ln(3)-\frac{0.02}{3}) $ so I agree with your answer.
My guess: The person creating the answers saw the "$\ln(3)$" and thought it to be "$\ln(1)$", which is zero. This would make the final result $-0.02 a/3$, which is answer (B).