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Step-by-Step Solution
Step 1: Understand the Given Conditions
We have a function $f$ that is:
Continuous on $[a,b]$.
Twice differentiable on $(a,b)$.
$f'(x) > 0$ for all $x \in (a,b)$, meaning $f$ is strictly increasing on $(a,b)$.
$f''(x) < 0$ for all $x \in (a,b)$, meaning the slope $f'(x)$ is strictly decreasing or equivalently that $f$ is concave down on $(a,b)$.
Step 2: Express the Differences Using the Mean Value Theorem
By the Mean Value Theorem (MVT), on the interval $[a,c]$, there exists some $\xi_1 \in (a,c)$ such that
$$
f(c) - f(a) \;=\; f'(\xi_1)\,(c - a).
$$
Similarly, on the interval $[c,b]$, there exists some $\xi_2 \in (c,b)$ such that
$$
f(b) - f(c) \;=\; f'(\xi_2)\,(b - c).
$$
Step 3: Compare the Values of $f'(\xi_1)$ and $f'(\xi_2)$
Because $f''(x) < 0$ on $(a,b)$, the derivative $f'(x)$ is strictly decreasing as $x$ increases. Thus if $\xi_1 < \xi_2$, we get:
$$
f'(\xi_1) \;>\; f'(\xi_2).
$$
Since $a < c < b$, we have $\xi_1 \in (a,c)$ and $\xi_2 \in (c,b)$. Consequently, $\xi_1 < \xi_2$ and therefore $f'(\xi_1) > f'(\xi_2)$.
Step 4: Form the Required Ratio
Using the MVT expressions, we write:
$$
\frac{f(c) - f(a)}{f(b) - f(c)}
\;=\; \frac{f'(\xi_1)\,(c - a)}{f'(\xi_2)\,(b - c)}.
$$
Since $f'(\xi_1) > f'(\xi_2)$ and both $c-a$ and $b-c$ are positive (because $a < c < b$), it follows that:
$$
\frac{f'(\xi_1)}{f'(\xi_2)}
\;>\; 1
\quad\Longrightarrow\quad
\frac{f(c) - f(a)}{f(b) - f(c)}
\;>\; \frac{c - a}{b - c}.
$$
Step 5: Conclude the Inequality
Hence, for any $c \in (a,b)$, we have shown:
$$
\frac{f(c) - f(a)}{f(b) - f(c)}
\;>\; \frac{c - a}{b - c}.
$$
This matches the answer given by Option 4.
Final Answer
The correct answer is
$$
\frac{c - a}{b - c}.
$$
Reference Diagram
