Prove this function has at most two zero points
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Let $f(x)$ be a twice differentiable function on the interval $(-infty,+infty)$ and $f''(x)>f(x)$ for all $x$. Prove that $f$ has at most two zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
edited Mar 9 at 19:17
Prem
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be a twice differentiable function on the interval $(-infty,+infty)$ and $f''(x)>f(x)$ for all $x$. Prove that $f$ has at most two zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
edited Mar 9 at 19:17
Prem
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be a twice differentiable function on the interval $(-infty,+infty)$ and $f''(x)>f(x)$ for all $x$. Prove that $f$ has at most two zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
edited Mar 9 at 19:17
Prem
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
$endgroup$
Let $f(x)$ be a twice differentiable function on the interval $(-infty,+infty)$ and $f''(x)>f(x)$ for all $x$. Prove that $f$ has at most two zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
real-analysis calculus
edited Mar 9 at 19:17
Prem
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
edited Mar 9 at 19:17
Prem
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
edited Mar 9 at 19:17
Prem
1134
edited Mar 9 at 19:17
Prem
1134
edited Mar 9 at 19:17
Prem
1134
1134
asked Mar 9 at 14:05
MaxwellMaxwell
445
asked Mar 9 at 14:05
MaxwellMaxwell
445
asked Mar 9 at 14:05
MaxwellMaxwell
445
445
add a comment |
add a comment |
1 Answer
1
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If $f$ has more than 2 zeros, then so does $e^-xf(x)=g(x)$. Then by the Rolle's theorem, $$h(x)=e^xg'(x)=f'(x)-f(x)$$ should have at least 2 zeros. But then $e^xh(x)$ has at least 2 zeros, which contradicts $$[e^xh(x)]'=e^x(h'(x)+h(x))=e^x(f''(x)-f(x))>0.$$
answered Mar 9 at 14:23
NaoNao
1836
$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $f$ has more than 2 zeros, then so does $e^-xf(x)=g(x)$. Then by the Rolle's theorem, $$h(x)=e^xg'(x)=f'(x)-f(x)$$ should have at least 2 zeros. But then $e^xh(x)$ has at least 2 zeros, which contradicts $$[e^xh(x)]'=e^x(h'(x)+h(x))=e^x(f''(x)-f(x))>0.$$
answered Mar 9 at 14:23
NaoNao
1836
$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
add a comment |
$begingroup$
If $f$ has more than 2 zeros, then so does $e^-xf(x)=g(x)$. Then by the Rolle's theorem, $$h(x)=e^xg'(x)=f'(x)-f(x)$$ should have at least 2 zeros. But then $e^xh(x)$ has at least 2 zeros, which contradicts $$[e^xh(x)]'=e^x(h'(x)+h(x))=e^x(f''(x)-f(x))>0.$$
answered Mar 9 at 14:23
NaoNao
1836
$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
add a comment |
$begingroup$
If $f$ has more than 2 zeros, then so does $e^-xf(x)=g(x)$. Then by the Rolle's theorem, $$h(x)=e^xg'(x)=f'(x)-f(x)$$ should have at least 2 zeros. But then $e^xh(x)$ has at least 2 zeros, which contradicts $$[e^xh(x)]'=e^x(h'(x)+h(x))=e^x(f''(x)-f(x))>0.$$
answered Mar 9 at 14:23
NaoNao
1836
$endgroup$
If $f$ has more than 2 zeros, then so does $e^-xf(x)=g(x)$. Then by the Rolle's theorem, $$h(x)=e^xg'(x)=f'(x)-f(x)$$ should have at least 2 zeros. But then $e^xh(x)$ has at least 2 zeros, which contradicts $$[e^xh(x)]'=e^x(h'(x)+h(x))=e^x(f''(x)-f(x))>0.$$
answered Mar 9 at 14:23
NaoNao
1836
answered Mar 9 at 14:23
NaoNao
1836
answered Mar 9 at 14:23
NaoNao
1836
answered Mar 9 at 14:23
NaoNao
1836
1836
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
add a comment |
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
Mar 9 at 14:40
add a comment |
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