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

Solution

Maths sir · 4 steps

Apply MVT from -7 to each point separately; the slope cap of 2 bounds each value from above.

1. MVT up to 0. On [-7, 0] there is a c with f(0) - f(-7) = 7f'(c). Since f'(c) ≤ 2, this gives f(0) + 3 ≤ 14, so f(0) ≤ 11.
2. MVT up to -1. On [-7, -1] there is a d with f(-1) - f(-7) = 6f'(d) ≤ 12. So f(-1) + 3 ≤ 12, which gives f(-1) ≤ 9.

   f(-1) + f(0) ≤ 9 + 11 = 20

3. Check both ends. The line f(x) = 2x + 11 has slope 2 and f(-7) = -3. It gives f(-1) = 9 and f(0) = 11, so the sum 20 is actually reached. There is no lower limit: only f'(x) ≤ 2 is given, so f can fall as steeply as it likes. For example, f(x) = -kx - 7k - 3 makes the sum -15k - 6, which goes to -∞ as k grows.
4. So. f(-1) + f(0) can take any value up to and including 20, so it lies in (-∞, 20], option (b).

The idea to keep. A one-sided bound on f' gives a one-sided bound on f. Use MVT from the known point to each unknown point, and add the bounds only after finding each one.

Answer from the NTA final answer key. Solution by PrepWiser.