How prove this the equation $f'(x)+f(x)=0$ has at least one root in $(a,b)$

Question

Suppose $f$ is continous on $[a,b]$ and differentiable on the open interval $(a,b)$,and such $$f(a)+f(b)=0$$ show that: then the equation $f'(x)+f(x)=0$ has at least one root in $(a,b)$

My try: not $$(e^xf(x))'=e^x(f(x)+f'(x))$$

But I can't,Thank you

Let $f(x) = 1 - x$ then $f(0) + f(2) = 0$, but $f'(x) + f(x) = -x$ and this is non zero on $(0, 2)$ — Paramanand Singh, Nov 06 '13 at 16:18
It is strange that people don't even verify the question and start rating it. — Paramanand Singh, Nov 06 '13 at 16:20
Are there counterexamples also if the interval for the roots includes $a$ and $b$? — Daniel R, Nov 07 '13 at 07:54

score 8 · Accepted Answer · answered Nov 06 '13 at 16:26

8

Please check your question. I believe there is a mistake. If $f(x) = 1 - x$ then $f(0) + f(2) = 0$ and $f'(x) + f(x) = -x$ which is non-zero in $(0, 2)$ so this serves as a counterexample. Please check the source of the question and re-write it.

answered Nov 06 '13 at 16:26

Paramanand Singh

87,309

Are there counterexamples also if the interval for the roots includes $a$ and $b$? – Daniel R Nov 07 '13 at 07:55
1

yes Daniel the same example in interval $[1/2, 3/2]$. You should have guessed this. – Paramanand Singh Nov 07 '13 at 10:05
1

Thanks for your answer. I don't care much for the attitude though, but never mind... – Daniel R Nov 07 '13 at 10:24

How prove this the equation $f'(x)+f(x)=0$ has at least one root in $(a,b)$

1 Answers1