Calculus problem (1)

Calculus Level 3

Given that $f(x+y) = f(x)+f(y)+2xy-1$ for all real $x$ and $y$ , where $f(x)$ is differentiable for all $x$ and that $f'(0) = \cos p$ .

How many of the following are correct?

$f'(1) = 2 + \cos p$
$f(1) = 2 + \cos p$
$f(0) = 0$
$f'(x) = 2x + x\cos p$
$f(x) = x(2x + x\cos p)$

4 1 2 5 0 3

1 solution

Tom Engelsman
Jun 30, 2017

If $f(x)$ is differentiable everywhere on its domain, then differentiating the above functional equation with respect to x and y each gives:

$f'(x+y) = f'(x) + 2y$

$f'(x+y) = f'(y) + 2x$

which after equating these two relationships gives the ODE $f'(x) = 2x +(f'(y) - 2y) = 2x + A$ . If $f'(0) = cos(p)$ , then $A = cos(p).$ Integrating this differential equation now yields:

$f(x) = x^2 + [cos(p)]x + B$ .

Now if we take $x = y = 0$ into our original functional equation, we acquire another boundary condition:

$f(0+0) = f(0) + f(0) + 2(0)(0) - 1 \Rightarrow f(0) = 1$

which ultimately results in $B = 1$ and a final solution of $\boxed{f(x) = x^2 + [cos(p)]x + 1}.$

Only choices (1) and (2) are correct.

Calculus problem (1)

1 solution

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