# Ap Calculus Solutions Free Response

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**Solution for the free response of AP Calculus AB! help ..?**

*1:. Given a function f with the following properties (i) f (x + h) = EHF* (X) + EXF (h) for all real numbers x and h. (Ii) f (x) is a derivative of all real numbers x (iii) f '(0) = 2. (A) Show f (0) = 0. (B) Using the definition of f '(0), find limits as Xapproaches 0 f (x) / x lim f (x) / x. (C) Show that there are a number real p such that f (x) = f (x) + x ^ eh for all real numbers x. (D) What is the value of the number p is described in (c)?

[This was very unpleasant, took me 3 hours to partially solve … if you just want the answer without all the wrong turns, scroll to the bottom] Given a function f with the following properties: (I) f (x + h) = EHF (x) + EXF (h) for all real numbers x and h. One might expect that knowledge of f (x + h) will allow us to calculate the derivative f '(X) of the definition of limit (assuming that f is differentiable). This should then indirectly, us f (x) when we integrate (do not forget the integration constant). But this seems to pop: f '(x) = h → 0 Lim (f (x + h) – f (x)) / h = lim h → 0 (EHF (x) + EXF (h)-f (X)) / h = lim h → 0 ((eh-1) f (x) + EXF (h)) / h So play with the definition to get a handle on it. It seems like e ^ x and / or logs will participate, and the function is homogeneous XYH (which are treated the same.) Setting x =- h gives us f (h + h) = EHF (-h) + e (-h) f (h) f (0) Eh = [f (H)-f (h)] Setting x = 0 is obtained: f (x 0) = e (0) f (x) + EXF (0) f (x) = EXF (0) f (x) / f (0) = setting ex x = h + gives (h + h) = EHF (h) + EHF (h) f (2h) 2eh = f (h) f (2h) / m (h) = 2eh Taking the log records … f (f (2h) / m (h)) = log (2 hours) x = 2 Framework h gives us f (2h + h) = EHF (2h) + e (2h) f (h) f (3h) = eh [f (2h) + 2f (h)] Substituting f (2h) = 2eh f (h ) … f (3h) = 2eh [eh f (h) + 2f (h)] f (3h) 2eh = f (h) [eh + 1] f (3h) / m (h) = (2eh) (eh + 1) Setting x = 3 h gives us f (3h + h ) eh = [3EH f (h) + 3f (h)] f (4h) = [f eh (3h) + 3f (h)] f (4h) / m (h) = (eh) ² (2 (eh + 1) + 3) f (4h) / m (h) = (eh) ² (2 (eh + 1) + 3) Tricky pattern … these do not help much:. * In addition, we will try to implicitly differentiate the best expression of the above to differentiate implicitly is no doubt: f (x) = EXF (0) f '(x) = EF (0) = k, we call e * f (0) = k (constant) then f (x) = kx, but then engage f (0) = K0 = 0, which in turn would imply f (x) ≡ 0 everywhere, which is a trivial solution. must have been something wrong with allowing both k * e = f (0), the which means that f (0) should be indefinite or infinite. Therefore the differentiation implicit registration of the form f (x) = EXF (0) log f (x) = 1 + log x + log f (0) 1 / f (x) * f '(x) = 1 / xf (x) = xf' (x) [*] f '(x) = f (x) / xf (x) = xf '(x) is very revealing, but we do not yet "t is f (x) in a closed … Then I hit this: f (x) = EXF (0) => f (x) / f (0) = ex This is much more manageable, so we're going to settle for the simple function g (x) = f (x) / f (0) g (x) = Ex, but keep in mind that g (0) = f ( 0) / F (0), which should be 1, not 0 as the former would So this suggests that f (x) is an example of something, we call h (x). f (x) = e ^ h (x) the second piece of information was as follows: f (x) = xf '(x) = e> ^ h (x) = e ^ h (x) h' (x) which means that h '(x) = 1 => h (x ) = x ² / 2 + k1 => f (x) e ^ = (x ² / 2 + k1) => f (x) K = e ^ (x ² / 2) Now check this satisfies the property f ( x + h) = EHF (x) + EXF (h) for all real numbers XYH: f (x + h) = K e ^ ((x + h) ² / 2) K = e ^ (x ² / 2). K e ^ (h ² / 2). K e ^ (2xh / 2) fails, but it is getting closer, and fulfills the f '(x) = xf (x) the requirement … (Ii) f (x) is a derivative of all real numbers x. Show that the above limit exists for all x, including 0 … Seems difficult. (Iii) f '(0) = 2 Where are derived from (i) and (Ii).

**u CAN learn AP Calculus! Chapter 6-1 Area Between Curves FRQ Free Response Question**