Suppose a differentiable function f(x) satisfies the identity
f(x+y) = f(x) + f(y) + xy² + x²y, for all real x and y.
lim _(x → 0) f( x )/x = 1, then f'(3) is equal to ______.

Solution

Maths sir · 3 steps

Do not hunt for f itself: put y = h in the identity and the derivative definition does the work.

1. Set up the derivative. By definition, f'(x) = lim(h → 0) (f(x+h) - f(x))/h. Put y = h in the identity to get f(x+h) - f(x) = f(h) + xh² + x² h.

   f'(x) = lim(h → 0) ( f(h)/h + xh + x² )

2. Take the limit. The given condition makes f(h)/h → 1, and xh → 0. So f'(x) = 1 + x² for every real x.
3. So. f'(3) = 1 + 3² = 1 + 9 = 10.

The common slip: 9. Dropping the lim(h → 0) f(h)/h = 1 term leaves f'(x) = x² and gives 9, not 10.

The idea to keep. For a functional equation in f(x+y), put y = h and use the first-principles derivative. The given limit at 0 is exactly the piece you need to finish the limit.

Answer from the NTA final answer key. Solution by PrepWiser.