Let f(x) be differentiable on [1, 6] and f(1) | vedclass

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(D) Given that $f(x)$ is differentiable on $[1, 6]$ and $f(1) = -2$. Since $f(x)$ has exactly one root in $(1, 6)$, let this root be $x_0$. For $f(x)$

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$f^{\prime}(c) > \frac{2}{5}$

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(D) Given that $f(x)$ is differentiable on $[1, 6]$ and $f(1) = -2$. Since $f(x)$ has exactly one root in $(1, 6)$, let this root be $x_0$. For $f(x)$ to have a root in $(1, 6)$, the function must change sign. Since $f(1) = -2  0$. By the Lagrange Mean Value Theorem $(LMVT)$, there exists at least one $c \in (1, 6)$ such that $f^{\prime}(c) = \frac{f(6) - f(1)}{6 - 1}$. Substituting the values, we get $f^{\prime}(c) = \frac{f(6) - (-2)}{5} = \frac{f(6) + 2}{5}$. Since $f(6) > 0$, it follows that $f(6) + 2 > 2$. Therefore, $f^{\prime}(c) = \frac{f(6) + 2}{5} > \frac{2}{5}$. Thus, there exists $c \in (1, 6)$ such that $f^{\prime}(c) > \frac{2}{5}$.

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 $c \in (1, 6)$ such that:

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