Question

Let f be differentiable for all x, if f(1) = -2 and f′(x) ≥ 2 for all
x∈[1,6], then the minimum value of f(6) is

Answer 1

Answer:

8

Step-by-step explanation:

by co-ordinate geometry...

x1 = 1 ; x2 = 6 & y1 = -2 ; y2 = ? ; m = f'(x) ≥ 2

since . .

y - y1 = m ( x - x1 )

y - (-2) ≥ 2 ( x - 1)

y + 2 ≥ 2x - 2

y ≥ 2x - 4

now......

y2 ≥ 2 (6) - 4

y2 ≥ 8

hence....

minimum value of y = 8