Let f be any function continuous on [a, b] and twice differentiable on (a, b). If for all x ∈ (a, b)

Q: Why can't we just use the definition of concavity directly?

You absolutely can, but using the Mean Value Theorem provides a rigorous, step-by-step bridge that connects the geometry of the curve to the algebra of the slopes. It turns a visual intuition into a bulletproof mathematical proof.

Q: Wait, why do we reject the negative root here?

We aren't rejecting a root, but we are being careful with inequality signs! When we cross-multiply, we must ensure the terms are positive. Since $f(x)$ is strictly increasing, $f(b) - f(c)$ is positive, and $c - a$ is positive. This allows us to maintain the direction of the inequality.

Q: Is this true for all functions?

No! This specific relationship holds only for functions that are strictly increasing and concave down. If the function were concave up, the inequality would flip. Always check your concavity!

Question

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$. If for all $x \in (a, b)$, $f'(x) > 0$ and $f''(x) < 0$, then for any $c \in (a, b)$, $\frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than:

Options:

(A) $\frac{b-c}{c-a}$

(B) $1$

(C) $\frac{c-a}{b-c}$

(D) $\frac{b+a}{b-a}$

Sigma Percentile · Accepted Answer

### The Geometry of the Curve Imagine you are standing on a path that is constantly rising, but as you walk forward, the path begins to level off. You are climbing, but the climb is getting easier. This is the physical reality of the function $f(x)$ described in our problem. We are given that $f'(x) > 0$, which tells us the function is strictly increasing—you are always gaining altitude. But then we have the condition $f''(x) < 0$. This is the signature of a concave down function, an upside-down bowl, meaning the rate at which you are climbing is slowing down. ### The Bridge of the Mean Value Theorem To turn this intuition into a rigorous proof, we need a bridge. We have an interval $[a, b]$ and a point $c$ in between, creating two sub-intervals: $[a, c]$ and $[c, b]$. Let's consider the chords connecting points $A(a, f(a))$, $C(c, f(c))$, and $B(b, f(b))$. The slopes of these chords are given by: $$m_1 = \frac{f(c) - f(a)}{c - a}$$ $$m_2 = \frac{f(b) - f(c)}{b - c}$$ Now, we invoke the Lagrange Mean Value Theorem (LMVT). It tells us that for the interval $[a, c]$, there exists a point $p \in (a, c)$ such that $f'(p) = m_1$. Similarly, for the interval $[c, b]$, there exists a point $q \in (c, b)$ such that $f'(q) = m_2$. ### The Inequality Dance Here is where the logic tightens. Since $p \in (a, c)$ and $q \in (c, b)$, it is undeniable that $p < c < q$, which implies $p < q$. Recall our condition $f''(x) < 0$. This implies that the first derivative $f'(x)$ is a strictly decreasing function. Therefore, if $p < q$, then the slope at $p$ must be greater than the slope at $q$: $$f'(p) > f'(q)$$ ### The Final Reveal We substitute our LMVT expressions back into the inequality: $$\frac{f(c) - f(a)}{c - a} > \frac{f(b) - f(c)}{b - c}$$ Because $f(x)$ is strictly increasing, $f(b) - f(c) > 0$, and because $c > a$, $c - a > 0$. We can safely cross-multiply these positive terms without flipping the inequality sign. Rearranging the terms, we arrive at the final conclusion: $$\frac{f(c) - f(a)}{f(b) - f(c)} > \frac{c - a}{b - c}$$ This is the mathematical proof that for a concave down function, the slopes of consecutive chords must decrease as you move from left to right. You have successfully mastered the interplay between geometry and calculus!

Animated Solution for Mathematics - Differentiation: Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

Select Answer:

Visualized Solution

Analyzing the Function's Behavior

Marking the Intervals

Constructing the Chords

Applying LMVT on $[a, c]$

Applying LMVT on $[c, b]$

Ordering the Points

Utilizing the Concavity Condition

Substituting the Slopes

Rearranging the Inequality

Final Conclusion

The Sigma Insight: Mean Value Theorems

The Geometry of the Curve

The Bridge of the Mean Value Theorem

The Inequality Dance

The Final Reveal

Similar Questions

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = f (1) = 1$ , then

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

Let $S$ be the set of all function $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ depending on $f$ , such that :

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

For all $x$ in $[0, 1]$ , let the second derivative $f^{''} (x)$ of a function $f (x)$ exist and satisfy $∣ f^{''} (x) ∣ < 1$ . If $f (0) = f (1)$ , then show that $∣ f^{'} (x) ∣ < 1$ for all $x$ in $[0, 1]$ .

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :