Let f : R R be a twice differentiable function such that the quadratic equation f(x)m2 - 2f'(x)m + f''(x) = 0 in m, has two equal roots for every x R. If f(0) = 1, f'(0) = 2, and ( , ) is the largest interval in which the function f( e x - x) is increasing, then + is equal to .
Answer
1
Solution
Equal roots discriminant = 0:
4[f'(x)]2 - 4f(x)f''(x) = 0 f(x)f''(x) - [f'(x)]2 = 0
f(x)f''(x) - [f'(x)]2[f(x)]2 = 0 ddx ( f'(x)f(x) ) = 0
f'(x)f(x) = k (constant); at x = 0: k = f'(0)f(0) = 2
ddx e f(x) = 2 e f(x) = 2x + C
f(0) = 1 C = 0 f(x) = e2x
Let g(x) = f( e x - x) = e2 e x - 2x = x2e-2x, x > 0
g'(x) = 2xe-2x - 2x2e-2x = 2x(1 - x)e-2x
For x > 0: g'(x) > 0 0 < x < 1
( , ) = (0, 1)
+ = 0 + 1 = 1
Answer: 1