Given that f(150)=100, f'(150)= -2.5, approximate the value of f(151.5).

The answer is approximately 96.25, but I'm unsure of how my teacher has gotten that answer.

Can someone please explain this to me?

1 answer

the 1st derivative is the slope of the tangent line , you can use it for approximation of a line from 150 to 151.5

(151.5 - 150) * -2.5 = -3.75

100 - 3.75 = 96.25