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| Question 1019853:  A function f is defined for all real numbers and has the following properties.
 • f(1) = 2
 • f(3) = 10
 • f'(2)=4
 • For all real numbers a and h, f(a+h)−f(a)=kah+nh^2−2nh (where k and n are
 constants).
 a) Find the value of k for this function.
 b) Find the value of n for this function.
 c) Find a formula for f'(x) (this should just be in terms of x, it should not depend on n or k).
 d) Using your answer to part c, find a formula for f(x) (this should just be in terms of x, it should not depend on n or k). Hint: Find a function g(x) such that g'(x) = f'(x).Then there is a theorem that there exists a (unique) real number c such that f(x) = g(x) + c.
 
 How do I isolate k and n so that I'm able to find out what they equal? I'm 100% stuck on this problem, and I'm not even sure where to start at any help would be appreciated, thank you for your time.
 Answer by robertb(5830)
      (Show Source): 
You can put this solution on YOUR website! We start with the difference f(3) - f(2) = k*2*1 +n -2n = 2k - n = 6, where we let a = 2 and h = 1. ==> 2k - n = 6   (Equation A)
 Similarly, f(2) - f(1) = k*1*1 +n -2n = k - n = 2, where we let a = 1 and h = 1.
 ==> k-n = 2    (Equation B)
 Solving for k and n from Equations A and B yields k = 4 and n = 2.
 With these values in mind, we get
 
  , where we replaced a by x. ==>
  , after dividing both sides by h. ==> f'(x) =
  Now f'(x) = 4x - 4 ==>
  for some constant c. We can use any one of the initial values given above to find the value of c.  In particular, f(1) = 2, and so
 
  , which gives c = 4. Therefore, f(x) =
  . Problem solved.
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