Prove this function has at most two zero points
$begingroup$
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 zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
asked 52 mins ago
MaxwellMaxwell
364
$endgroup$
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 zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
asked 52 mins ago
MaxwellMaxwell
364
$endgroup$
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 zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
asked 52 mins ago
MaxwellMaxwell
364
$endgroup$
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 zero points.
I'm trying to prove it by contradiction,but I can't work it out.
real-analysis calculus
real-analysis calculus
asked 52 mins ago
MaxwellMaxwell
364
asked 52 mins ago
MaxwellMaxwell
364
asked 52 mins ago
MaxwellMaxwell
364
asked 52 mins ago
MaxwellMaxwell
364
asked 52 mins ago
MaxwellMaxwell
364
364
add a comment |
add a comment |
1 Answer
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$begingroup$
If $f$ has more than 2 zeros, then so does $e^{-x}f(x)=g(x)$. 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 34 mins ago
NaoNao
713
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Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
If $f$ has more than 2 zeros, then so does $e^{-x}f(x)=g(x)$. 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 34 mins ago
NaoNao
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
add a comment |
$begingroup$
If $f$ has more than 2 zeros, then so does $e^{-x}f(x)=g(x)$. 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 34 mins ago
NaoNao
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
add a comment |
$begingroup$
If $f$ has more than 2 zeros, then so does $e^{-x}f(x)=g(x)$. 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 34 mins ago
NaoNao
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $f$ has more than 2 zeros, then so does $e^{-x}f(x)=g(x)$. 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 34 mins ago
NaoNao
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 34 mins ago
NaoNao
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 34 mins ago
NaoNao
713
answered 34 mins ago
NaoNao
713
713
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nao is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
add a comment |
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
$begingroup$
Thank you so much!tailihaile!
$endgroup$
– Maxwell
17 mins ago
add a comment |
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