Dave uses the tangent line approximation on a certain function f(x) and obtains the approximations f(2.98) is about 2.08 and f(3.01) is about 1.96. What does the tangent line approximation give for f(2.99)?
+37159 I suppose you are learning interpolation .... a type of extimation that assumes the graph is linear between two points (2.98 and 3.01 in this case)
from (f(2.98) to f (3.01) is 1.96 - 2.08 = - .12
(3.01 - 2.98) is .03
-.12/.03 = -4 the graph changes -4 per 1 change in f(x)
for .01 change (from 2.98 to 2.99)
it will be 2.08 - 4(.01) = 2.04
