Question

If \(F\) is a differentiable function such that \(F(3) = 6\) and \(F(9) = 2\), then there must exist at least one number 'a' between 3 and 9, such that:

• A. \(F'(a) = \dfrac{3}{2}\)
• B. \(F(a) = -\dfrac{3}{2}\)
• C. \(F'(a) = -\dfrac{3}{2}\)
• D. \(F'(a) = -\dfrac{2}{3}\)

Hint:

Use the Lagrange Mean Value Theorem: the derivative at some point between 3 and 9 must equal the average rate of change (F(9)-F(3))/(9-3).

Answer 1

The Correct Option is D

Step 1: Identify which theorem applies.
We are told \(F\) is differentiable on an interval containing \([3,9]\), and we know its values at the two endpoints, \(F(3)=6\) and \(F(9)=2\). Whenever a function is continuous on a closed interval and differentiable on the open interval, the Lagrange Mean Value Theorem (LMVT) guarantees that somewhere strictly between the endpoints, the instantaneous rate of change equals the average rate of change over the whole interval.

Step 2: State the theorem precisely.
LMVT says there exists at least one point \(a\) with \(3 < a < 9\) such that
\[ F'(a) = \frac{F(9)-F(3)}{9-3} \] This is just the slope of the line joining the two endpoints of the graph of \(F\).

Step 3: Plug in the given values.
\[ F'(a) = \frac{2-6}{9-3} = \frac{-4}{6} = -\frac{2}{3} \]
Step 4: Check why the other options are wrong.
Option (A) has the wrong sign; since \(F\) decreases overall from 6 to 2, its average rate of change is negative, ruling out a positive \(\frac{3}{2}\). Option (B) talks about \(F(a)\), the value of the function itself, not its derivative; LMVT makes a statement about the derivative, not the function's value, so this option confuses the two. Option (C), \(-\frac{3}{2}\), looks like a plausible distractor but does not match the actual computed average rate of \(-\frac{4}{6}\).

Final Answer:
By the Mean Value Theorem, there is a point \(a\) in \((3,9)\) where the derivative equals the average rate of change. \[ \boxed{F'(a) = -\frac{2}{3}} \]