## Calculus 8th Edition

$15.968$
The function $x^4$ is seen here at $a = 1.999$ We can approximate the slope at 2 to be... $$f'(x) = 4x^3$$ $$f'(2) = 32$$ Since the slope is 32, Since the x value decreased by $0.001$ the y value decreased by about $0.001 * 32$ $$f(1.999) = f(2) - 0.032 = 16 - 0.032 = \boxed{15.968}$$