Question

Let $f(x)=x^{2}-10x+16$, $x\in\mathbb{R}.$ If $f^{\prime}(c)$ is equal to slope of the straight line joining the points (2,0) and (8,0), then the value of $c$ is

• A. 1
• B. 2
• C. 4
• D. 5
• E. 6

Hint:

Logic Tip: A powerful shortcut for ANY quadratic function $f(x) = ax^2+bx+c$: The point $c$ where the tangent is parallel to the secant line between $x_1$ and $x_2$ is EXACTLY the midpoint. $c = \frac{x_1 + x_2}{2}$. Here, $\frac{2 + 8}{2} = 5$. No calculus required!

Answer 1

The Correct Option is D

Concept:
This problem applies Lagrange's Mean Value Theorem (LMVT). LMVT states that there exists a point $c$ in $(a, b)$ such that the instantaneous rate of change (derivative) at $c$ equals the average rate of change (secant slope) over $[a, b]$: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$
Step 1: Calculate the slope of the secant line.
We are given two points: $(2, 0)$ and $(8, 0)$. The slope $m$ of the line joining these points is: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{8 - 2} = \frac{0}{6} = 0$$
Step 2: Find the derivative of the function.
Given $f(x) = x^2 - 10x + 16$. $$f'(x) = 2x - 10$$ Therefore, the derivative at point $c$ is: $$f'(c) = 2c - 10$$
Step 3: Equate the derivative to the secant slope and solve for c.
The problem states that $f'(c)$ equals the slope $m$: $$2c - 10 = 0$$ $$2c = 10$$ $$c = 5$$