Let $f(x)=\dfrac{20}{(x^6+x^4+x^2+1)}$.
I need to show that for any $k \in [5,20]$, there is a point $c \in [0,1]$ such that $f(c)=k$.
Thanks in advance.
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1Do you know the intermediate value theorem? – Chris Eagle Apr 02 '13 at 18:28
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Start by computing $f(0)$ and $f(1)$. – Julien Apr 02 '13 at 18:29
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Yeah, in essence, if $f$ is cts on $[a,b]$ and and $k$ is a number between $f(a)$ and $f(b)$, then there is some point $c \in (a,b)$ s.t. $f(c)=k$ – Peej Gerard Apr 02 '13 at 18:30
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hmm so $f(0)= 20$ and $f(1)=5$ – Peej Gerard Apr 02 '13 at 18:31
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@PaulGerard Now, put $a=0$, $b=1$ to what you have written. – Lord Soth Apr 02 '13 at 18:31
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Is it that easy to simply plug in my values for $a$ and $b$ and use the IVT to justify? – Peej Gerard Apr 02 '13 at 18:32
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Yes, it is, provided that you can show $f(x)$ is continuous on $[0,1]$. – Lord Soth Apr 02 '13 at 18:33
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Can i similarly show that for any $k \in (0,5]$, there is a point $c \geq 1$ s.t. $f(c)=k$? – Peej Gerard Apr 02 '13 at 18:37
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May be? If you can prove that there exists a point $d\ge1$ such that $f(d)<k$ and $f$ is continuout also in $[1,d]$. – Jyrki Lahtonen Apr 02 '13 at 18:41
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Observe that $x^6+x^4+x^2+1>0 \quad \forall x\in \mathbb{R}$ and is continuous(see proving continuity of polynomials). Now $f(x)$ is continuous on $\mathbb{R}$ and observe that $f(1)=5, f(0)=20$. Intermediate value theorem tells that for every $k$ such that $5\le k\le 20$ there exists a $c\in [0,1]$ such that $f(c)=k$.
