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Let $f(x)$ be differentiable on $[1,6]$ and $f(1)=-2$. If $f(x)$ has only one root in $(1,6)$ then there exists $\mathcal{C} \in(1,6)$ such that
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Verified Answer
The correct answer is:
$f^{\prime}(\mathrm{c})>\frac{2}{5}$
$f(x)$ is differentiable on $[1,6]$
$$
f(1)=-2
$$
$f(x)$ has only are roots in $(1,6)$
$$
\because
$$
$f(1) f(6) < 0$
(-2) $f(6) < 0 \Rightarrow f(6)>0$
By L.M.V.T. $\quad f^{\prime}(c)=\frac{f(6)-f(1)}{6-1} \Rightarrow f^{\prime}(c)=\frac{f(6)+2}{5}$
$$
\Rightarrow \quad f^{\prime}(c)>\frac{2}{5} \quad[\because f(6)>0]
$$
$$
f(1)=-2
$$
$f(x)$ has only are roots in $(1,6)$
$$
\because
$$
$f(1) f(6) < 0$
(-2) $f(6) < 0 \Rightarrow f(6)>0$
By L.M.V.T. $\quad f^{\prime}(c)=\frac{f(6)-f(1)}{6-1} \Rightarrow f^{\prime}(c)=\frac{f(6)+2}{5}$
$$
\Rightarrow \quad f^{\prime}(c)>\frac{2}{5} \quad[\because f(6)>0]
$$
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