Question

If a function \(f:[0,1]\to\mathbb{R}\) is differentiable, \(f'(0) = -1\) and \(f'(1) = 5\), then which one of the following statements is true?

• A. There exists a point \(c\in(0,1)\) such that \(f'(c) = -1\).
• B. f' must be continuous on [0,1].
• C. There exists a point \(c\in(0,1)\) such that \(f'(c) = 4\).
• D. f' must be differentiable on (0,1).

Hint:

Apply Darboux's theorem, which gives derivatives the intermediate value property even when the derivative is not continuous.

Answer 1

The Correct Option is C

Step 1: Prove the needed instance of Darboux's theorem directly, without citing it as a black box. Define $g(x) = f(x) - 4x$ on $[0,1]$. Then $g$ is differentiable on $[0,1]$ with $g'(x) = f'(x) - 4$.

Step 2: Compute g' at the endpoints. $g'(0) = f'(0) - 4 = -1-4 = -5 < 0$ and $g'(1) = f'(1) - 4 = 5-4 = 1 > 0$.

Step 3: Use the definition of derivative to locate a local minimum of g in the interior. Since $g'(0) < 0$, $g$ is decreasing at $0$, so $g(x) < g(0)$ for $x$ slightly greater than $0$. Since $g'(1) > 0$, $g$ is increasing at $1$, so $g(x) < g(1)$ for $x$ slightly less than $1$. Thus $g$ does not attain its minimum on $[0,1]$ at either endpoint.

Step 4: Apply the extreme value theorem. Since $g$ is differentiable, it is continuous on the closed bounded interval $[0,1]$, so by the extreme value theorem $g$ attains a global minimum at some point $c \in [0,1]$. By Step 3, $c$ cannot be $0$ or $1$, so $c \in (0,1)$, an interior point.

Step 5: Apply Fermat's theorem on interior extrema. Since $g$ is differentiable at the interior minimum $c$, Fermat's theorem gives $g'(c) = 0$, i.e. $f'(c) - 4 = 0$, so $f'(c) = 4$.

Step 6: Conclude. This directly exhibits the required point $c \in (0,1)$ with $f'(c)=4$, confirming option (C) by an explicit construction rather than by quoting Darboux's theorem. \[\boxed{f'(c) = 4 \text{ for some } c \in (0,1)}\]