# Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?

• Apr 30th 2009, 07:29 AM
fardeen_gen
Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?
If $m^2 + n^2 - mn - m - n - 1\leq 0$, where $m,n\in\mathbb{R}$ and $z_{1},z_{2}$ be the complex numbers such that that $|z_{1}| \leq m, |z_{2}|\leq n$ and $a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2$. If $f(x) = ax^2 + bx + c > 0$ for all $x$, show that $f(x) + f'(x) + f''(x) > 0$ for all real $x$.
• Apr 30th 2009, 01:56 PM
Opalg
Quote:

Originally Posted by fardeen_gen
If $\color{blue}m^2 + n^2 - mn - m - n - 1\leq 0$, where $\color{blue}m,n\in\mathbb{R}$ and $\color{blue}z_{1},z_{2}$ be the complex numbers such that that $\color{blue}|z_{1}| \leq m, |z_{2}|\leq n$ and $\color{blue}a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2$. If $f(x) = ax^2 + bx + c > 0$ for all $x$, show that $f(x) + f'(x) + f''(x) > 0$ for all real $x$.

You have mixed up two separate problems. The sentence in blue is completely irrelevant to the problem contained in the second sentence.

The conditions for the quadratic polynomial $f(x) = ax^2 + bx + c$ to be always positive are that $a>0$ and $b^2<4ac$. If f(x) satisfies those conditions then it easy to check that $f(x) + f'(x) + f''(x) = ax^2 + (2a+b)x + (2a+b+c)$ satisfies the same conditions.