Question

Let $f(x)=\sqrt{x}$, $4\le x\le16$. If the point $c\in(4,16)$ is such that the tangent line to the graph of $f$ at $x=c$ is parallel to the chord joining $(16,4)$ and $(4,2)$, then the value of $c$ is

• A. 7
• B. 9
• C. 10
• D. 11
• E. 14

Hint:

Calculus Tip: The Mean Value Theorem guarantees that if you draw a straight line between any two points on a smooth curve, there is always at least one point in between where the curve's tangent is perfectly parallel to that line.

Answer 1

The Correct Option is B

Concept:
This problem is a direct application of the Mean Value Theorem (MVT), which states that for a continuous and differentiable function, there exists some point $c$ in the interval $(a, b)$ where the instantaneous rate of change (the derivative $f^{\prime}(c)$) equals the average rate of change (the slope of the chord connecting the endpoints).

Step 1: Calculate the slope of the chord.
Use the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ for the points $(4, 2)$ and $(16, 4)$: $$m_{chord} = \frac{4 - 2}{16 - 4}$$ $$m_{chord} = \frac{2}{12} = \frac{1}{6}$$

Step 2: Find the derivative of the function.
Differentiate $f(x) = \sqrt{x} = x^{1/2}$ with respect to $x$: $$f^{\prime}(x) = \frac{1}{2}x^{-1/2}$$ $$f^{\prime}(x) = \frac{1}{2\sqrt{x}}$$

Step 3: Set the derivative equal to the chord slope.
Since the tangent line at $x = c$ is parallel to the chord, their slopes must be identical. Substitute $x = c$ and set it equal to $1/6$: $$f^{\prime}(c) = \frac{1}{2\sqrt{c}} = \frac{1}{6}$$

Step 4: Solve the resulting equation for c.
Cross-multiply to isolate the square root term: $$2\sqrt{c} = 6$$ Divide both sides by 2: $$\sqrt{c} = 3$$

Step 5: Calculate the final value.
Square both sides of the equation to eliminate the radical and find $c$: $$c = 3^2 = 9$$ Check that $9$ is indeed within the open interval $(4, 16)$, which it is. Hence the correct answer is (B) 9.