Let $f$ be continuous on $[0,1]$ and such that $f(0)<0$ and $f(1)>1$. Prove that there exists $c\in(0,1)$ such that $f(c)=c^4$
Tell me more
×
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.
|
|
If we let $g(x)=f(x)-x^4$, then we are given $g(0)=f(0)-0<0$ and $g(1)=f(1)-1>0$. Since $f$ is continuous and $x^4$ is continuous, their difference is, and $g$ is defined on the interval $[0,1]$, so we can apply the Intermediate Value Theorem to $g$. In particular, since $g(0)<0<g(1)$, there exists $c\in (0,1)$ such that $g(c)=0$, so $f(c)-c^4=0$, or $f(c)=c^4$ as we desired. |
|||
|
|
|
consider $$g(x)=f(x)-x^4$$ $g(x)$ is continuous from $[0,1]$ since both $f(x)$ and $x^4$ continuous on $[0,1]$. Now, $g(0)=f(0)-0^4=a<0$ (since $f(0)<0$ is given) $g(1)=f(1)-1^4 = b>0$ (since $f(1)>1$ is given) $g(x)$ is continuous, so by Intermediate Value Theorem, for every $d$ between $g(0)$ and $g(1)$ $\exists\,c\in(0,1)$ such that $g(c)=d$ Choose $d=0$ so we have $g(c)=0$ or $g(c)=f(c)-c^4=0$ which implies $$f(c)=c^4\, for c\in(0,1)$$ |
|||
|
