Exercise 5.3.3

Question

Let $h$ be a differentiable function defined on the interval $[0,3]$, and assume that $h(0)=1, h(1)=2$, and $h(3)=2$.
  \begin{enumerate}[label=(\alph*)] 
  \item Argue that there exists a point $d \in[0,3]$ where $h(d)=d$.
  \item Argue that at some point $c$ we have $h^{\prime}(c)=1 / 3$.
  \item Argue that $h^{\prime}(x)=1 / 4$ at some point in the domain.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Consider $g(x) = h(x) - x$ which is continuous and has $g(0) = 1$ and $g(3) = -1$, then apply the IVT to find $d \in (0,3)$ with $g(d) = 0$ which implies $h(d) = d$.
  \item Apply MVT on $[0,3]$ to get $c \in (0,3)$ with
    $$h'(c) = \frac{h(0) - h(3)}{0 - 3} = 1/3$$
  \item We can find $c \in (0, 3)$ with $h'(c) = 1/3$ and a $d \in (1,3)$ with $h'(d) = 0$. So by Darboux's theorem there exists a point $x \in (c,d)$ with $h'(x) = 1/4$.
   \end{enumerate}

Exercise 5.3.3

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Exercise 5.3.3

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