Part 6/7:
And to approximate ( f'(1.05) ):
[
L(1.05) = \frac{1}{4}(1.05 - 1) + 2 = \frac{0.05}{4} + 2 = 2.0125
]
Accuracy of Approximations
Always remember that while linear approximations can provide a quick and relatively accurate estimate, they are best used close to the point of tangency. The difference between the predicted and actual values may become larger as we move further away from that point. In our examples, checking these with exact calculations showed that the linear estimates were reasonably close but not perfect.