Question

Let the function, ƒ:[-7, 0]\( \to \)R be continuous on [-7,0] and differentiable on (-7, 0). If ƒ(-7) = - 3 and ƒ'(x) \( \le \) 2, for all x \( \in \) (-7,0), then for all such functions ƒ, ƒ(-1) + ƒ(0) lies in the interval:

• A. \( \left[ { - 6,20} \right] \)
• B. \( \left( { - \infty ,\left. {20} \right]} \right. \)
• C. \( \left[ { - 3,11} \right] \)
• D. \( \left( { - \infty ,\left. {11} \right]} \right. \)

Hint:

Apply the Mean Value Theorem on [-7, -1] and [-7, 0] to bound f(-1) and f(0) from above; there is no lower bound.

Answer 1

The Correct Option is B

Step 1: Define \(g(x)=f(x)-2x\). Then \(g'(x)=f'(x)-2\le0\), so g is non-increasing on [-7, 0].
Step 2: \(g(-7)=-3+14=11\), so \(g(x)\le11\) for all x in [-7, 0].
Step 3: \(f(-1)+f(0)=g(-1)+g(0)+2(-1)+2(0)=g(-1)+g(0)-2\le22-2=20\).
Step 4: g can fall arbitrarily far (e.g. f(x) = -3 - M(x+7) for large M), so the sum has no lower bound.
Step 5: The set of values is \((-\infty,20]\).
Answer: (B)