Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would appreciate if somebody could help me with the following problem:

Q: If $f''(x)$ is exist, $f'(0)>0, f(0)>0, f''(x)>f(x)~(0\leq x\leq 1)$ then $f(x)>0(0\leq x\leq 1)$

share|cite|improve this question
up vote 3 down vote accepted

Assume there exists $x\in[0,1]$ such that $f(x)\leq 0$. By the intermediate value theorem, the compact set $$ \{x\in[0,1]\;;\;f(x)=0\} $$ is nonempty. Let $x_0$ be its minimum. Note that $x_0\in(0,1]$ and remember that $f(x_0)=0$. Then $f(x)>0$ for all $x\in[0,x_0)$. So $f''(x)>0$ for all $x\in[0,x_0)$. Hence $f'$ is increasing on $[0,x_0]$. Since $f'(0)>0$, we get $f'(x)>0$ on $[0,x_0]$. So $f$ is increasing on $[0,x_0]$. Since $f(0)>0$, we get $f(x_0)>0$. Contradiction.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.