Applications of Derivatives
Concave Functions and Extrema
Grade 12
Question:
<p>If $f''(x) < 0$ for all $x \in (a,b)$ and $c$ is a point such that $a < c < b$, and $(c, f(c))$ is the point lying on the curve for which $f(c)$ is maximum, then $f'(c)$ is equal to</p>
<p>(A) $\displaystyle \frac{f(b) - f(a)}{b-a}$</p>
<p>(B) $\displaystyle \frac{2(f(b) - f(a))}{b-a}$</p>
<p>(C) $\displaystyle \frac{2f(b) - f(a)}{2b-a}$</p>
<p>(D) 0</p>
Step-by-Step Solution
Key Concept: Use properties of concave functions and Mean Value Theorem: for a concave function with maximum at interior point, the derivative equals the secant slope.
<p>Since $f''(x) < 0$ for all $x \in (a,b)$, the function is concave on this interval. If $(c, f(c))$ is a point where $f$ attains its maximum on the interval, then by Lagrange's Mean Value Theorem applied to the concavity property and the fact that $f'(c) = 0$ would contradict the maximum being at an interior point for a concave function, we have:</p><p>By the Mean Value Theorem: $f'(c) = \displaystyle \frac{f(b) - f(a)}{b-a}$</p>
Correct Answer: A