Answer:
f(11) ≈ 52
f(9.5) ≈ 49
f(10.25) ≈ 50.5
Explanation:
Given a function value (f(10) = 50) and its derivative (f'(10) = 2), you want approximations of nearby points.
Linear approximation
The linear approximation of the function near x=10 is given by ...
f(x) ≈ f(10) +f'(10)(x -10)
f(x) = 50 +2(x -10) = 30 +2x
Application
f(11) ≈ 30 +2·11 = 52
f(9.5) ≈ 30 +2·9.5 = 49
f(10.25) ≈ 30 +2·10.25 = 50.50