The function f is twice differentiable with f(2)=1, f'(2)=4, and f''(2)=3. what is the value of the approximation of f(1.9) using the line tangent to the graph of f at x=2?

 The line tangent to the graph of y = f(x) at x = a has slope f'(a) and goes through the point (a, f(a)). So it has the equation 

(y - f(a))/(x - a) = f'(a) 

[sometimes called the 'point-slope' form of the line] which can be rewritten in perhaps the more familiar form 

y = f'(a) (x - a) + f(a). 

Plugging in a = 2 and using the given information, we see the tangent line to the graph of y = f(x) at x = 2 has equation 

y = f'(2) (x - 2) + f(2) = 4 (x - 2) + 1. 

To compute an "approximation of f(1.9) using the line tangent to the graph of f at x = 2" means to use the equation for the tangent line with x = 1.9 plugged into it as a substitute for f(1.9). When you plug 1.9 in here you get 

4 (1.9 - 2) + 1 = 4(-0.1) + 1 = -0.4 + 1 = 0.6 

which is the answer. So, I guess I don't know how they got 0.4 either; it's wrong. 

You may wonder what happened with f''(2) = 3. Well, it doesn't enter into it at all. (The tangent line to the graph, the only thing we need, is determined by f(2) and f'(2).) Presumably its value was given to test the understanding of people solving the problem--- if they somehow bring f''(2) into it, it shows that they are just trying random things involving the given information and don't get the problem. This may be useful if the grader wants to distinguish between different levels of wrongness when assigning partial credit.

The answer would be -3.9> -4.5